datavorous/entrance-exam-dataset huggingface dataset:
The ionization energy of gaseous Na atoms is 495.5 kJ mol-1. The lowest possible frequency of light that ionizes a sodium atom is

(h=6.626×10-34 Js, NA=6.022×1023 mol-1)

Options:
A 1.24×1015 s-1
B 7.50×104 s-1
C 4.76×1014 s-1
D 3.15×1015 s-1

The correct answer is:
1.24×1015 s-1

Solution:
One photon ionises one Na-atom by elastic collision.
NA hvphoton×10–3 (KJ mole–1)
6·022×1023×6·626×10-34×υphoton×10-3=495·5
υphoton=495·56·022×6·626×10-14
=12.41×1014
=1·241×1015 sec-1

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2014 (19 Apr Online)
The ionization energy of hydrogen is $13.6 \mathrm{eV}$. The energy of the photon released when an electron jumps from the first excited state $(n=2)$ to the ground state of a hydrogen atom is

Options:
A $3.4 \mathrm{eV}$
B $4.53 \mathrm{eV}$
C $10.2 \mathrm{eV}$
D $13.6 \mathrm{eV}$

The correct answer is:
$10.2 \mathrm{eV}$

Solution:
Energy required to removed electron in th $(n=2)$ stage The energy of the photon
$\begin{array}{l}
=R\left(\frac{1}{1^{2}}-\frac{1}{n^{2}}\right) \Rightarrow 13.6 \mathrm{eV}\left(\frac{1}{1}-\frac{1}{4}\right) \\
=13.6 \mathrm{eV}\left(\frac{4-1}{4}\right)=13.6 \mathrm{eV} \times \frac{3}{4}=\frac{40.8}{4} \mathrm{eV} \\
=10.2 \mathrm{eV}
\end{array}$

Topic: Physics, Atomic Physics, JEE Main
The ionization energy of nitrogen is more than that of oxygen because

Options:
A Nitrogen has half filled p-orbitals
B Nitrogen is left to the oxygen in the same period of the periodic table
C Nitrogen contains less number of electrons
D Nitrogen is less electronegative

The correct answer is:
Nitrogen has half filled p-orbitals

Solution:
Nitrogen has half filled p-orbitals

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The ionization energy of sodium in kJ mol–1 . If electromagnetic radiation of wavelength 242 nm is just sufficient to ionize sodium atom is ______.(nearest integer)

Options:

The correct answer is:
494

Solution:
The ionisation energy of one sodium atom =hcλ

& the ionisation energy of one mole Na atom =hcλNA=6.62×10-34×3×108×6.02×1023242×10-9=494.03 kJ/mol

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2024 (31 Jan Shift 1)
The ionization energy of the electron in the hydrogen atom in its grounds state is $13.6 \mathrm{eV}$. The atoms are excited to higher energy levels to emit radiations of 6 wavelengths. Maximum wavelength of emitted radiation corresponds to the transition between :

Options:
A $n=4$ to $n=3$ states
B $n=3$ to $n=2$ states
C $n=3$ to $n=1$ states
D $n=2$ to $n=1$ states

The correct answer is:
$n=4$ to $n=3$ states

Solution:
$\begin{aligned} & \frac{\mathrm{n}(\mathrm{n}-1)}{2}=6 \\ \Rightarrow & \mathrm{n}=4\end{aligned}$
For maximum wavelength energy difference between states should be minimum because
$$
\lambda=\frac{\mathrm{hc}}{\Delta \mathrm{E}}
$$
So, transition state in $\mathrm{n}=4$ to $\mathrm{n}=3$

Topic: Physics, Atomic Physics, NEET, NEET 2009 (Mains)
The ionization enthalpies of Li and Na are 520 kJ mol-1 and 495 kJ mol-1 respectively. The energy required to convert all the atoms present in 70 mg of Li vapours and 230 mg of sodium vapours of their respective gaseous cations respectively are

Options:
A 5200 J, 4950 J
B 4950 J, 5200 J
C 52 J, 49.5 J
D 495 J, 520 J

The correct answer is:
5200 J, 4950 J

Solution:
Number of moles =weight in gmatomic weight
Number of moles Li=701000Ã7=10-2
Number of moles of Na=2301000Ã23Ã=10-2
Amount of energy = moles Ã energy needed for 1 mole.
So the amounts of energies required for 10-2 mole each of Li and Na are as follows
For sodium =520Ã10-2 kJ=5200 J
For Lithium =495Ã10-2 kJ=4950 J respectively.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The ionization enthalpy of hydrogen atom is $1.312 \times 10^6 \mathrm{Jmol}^{-1}$. The energy required to excite the electron in the atom from $n=1$ to $n=2$ is

Options:
A
$8.51 \times 10^5 \mathrm{Jmol}^{-1}$

B
$6.56 \times 10^5 \mathrm{Jmol}^{-1}$

C
$7.56 \times 10^5 \mathrm{Jmol}^{-1}$

D
$9.84 \times 10^5 \mathrm{Jmol}^{-1}$

The correct answer is:
$9.84 \times 10^5 \mathrm{Jmol}^{-1}$

Solution:
$$
\begin{aligned}
& \Delta \mathrm{E}=\mathrm{E}_2-\mathrm{E}_1=-\frac{1.312 \times 10^6}{2^2}-\left(-\frac{1.312 \times 10^6}{1}\right) \\
& =9.84 \times 10^5 \mathrm{~J} \mathrm{~mol}^{-1}
\end{aligned}
$$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2008
The ionization enthalpy of Na+ formation from Nag is 495.8 kJ mol-1, while the electron gain enthalpy of Br is -325.0 kJ mol-1. Given the lattice enthalpy of NaBr is -728.4 kJ mol -1 . The energy for the formation of NaBr ionic solid is -_____10-1 kJ mol-1

Options:

The correct answer is:
5576

Solution:
ΔHformation =IE1+ΔHeg1+LE

=495.8+-325+-728.4

=-557.6

=-5576×10-1KJ/mol .

Note: The above calculation is not for

ΔHformation but for ΔHReaction

But on the basis of given data it is the best ans.

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2021 (25 Feb Shift 1)
The ionization isomer of $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}\left(\mathrm{NO}_2\right)\right] \mathrm{Cl}$ is

Options:
A
$\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4\left(\mathrm{O}_2 \mathrm{~N}\right)\right] \mathrm{Cl}_2$

B
$\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}_2\right]\left(\mathrm{NO}_2\right)$

C
$\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}(\mathrm{ONO})\right] \mathrm{Cl}$

D
$\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}_2\left(\mathrm{NO}_2\right)\right] \cdot \mathrm{H}_2 \mathrm{O}$

The correct answer is:
$\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}_2\right]\left(\mathrm{NO}_2\right)$

Solution:
The ionization isomer of $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}\left(\mathrm{NO}_2\right)\right] \mathrm{Cl}$ is $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}_2\right] \mathrm{NO}_2$ because of exchanging of ligand and counter ions
Coordination chemistry
Straight conceptual
II

Topic: Chemistry, Coordination Compounds, JEE Main
The ions $\mathrm{O}^{2-}, \mathrm{F}^{-}, \mathrm{Na}^{+}, \mathrm{Mg}^{2+}$ and $\mathrm{Al}^{3+}$ are isoelectronic. Their ionic radii show:

Options:
A a significant increase from $\mathrm{O}^{2-}$ to $\mathrm{Al}^{3+}$
B a significant decrease from $\mathrm{O}^{2-}$ to $\mathrm{Al}^{3+}$
C an increase from $\mathrm{O}^{2-}$ to $\mathrm{F}^{-}$and then decrease from $\mathrm{Na}^{+}$to $\mathrm{Al}^{3+}$
D a decrease from $\mathrm{O}^{2-}$ to $\mathrm{F}^{-}$and then increase from $\mathrm{Na}^{+}$to $\mathrm{Al}^{3+}$

The correct answer is:
a significant decrease from $\mathrm{O}^{2-}$ to $\mathrm{Al}^{3+}$

Solution:
Ionic radii decrease with the increase in charge on species.

Topic: Chemistry, Structure of Atom, JEE Main
The ions present in Al4C3,CaC2 and Mg2C3 are respectively

Options:
A C4-,C22-,C34-
B C22-,C4-,C34-
C C34-,C32-,C4-
D C34-,C4-,C22-

The correct answer is:
C4-,C22-,C34-

Solution:
C4-,C22-,C34- are present in Al4C3,CaC2 and Mg2C3 respectively.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
The isoelectronic set of ions is:

Options:
A F-, Li+, Na+ and Mg2+
B N3-, Li+, Mg2+ and O2-
C N3-, O2-, F- and Na+
D Li+, Na+, O2- and F-

The correct answer is:
N3-, O2-, F- and Na+

Solution:
In Isoelectronic species, number of electrons should be same
Number of electrons present in Li+= 3-1 = 2
Number of electrons present in Na+= 11-1 = 10
(A) F-=9+1=10e-
Li+=3-1=2e-
Na+=11-1=10e-
Mg+2=12-2=10e-
(B) N-3=7+3=10e-
Li+=3-1=2e-
Mg+2=12-2=10e-
Na+=11-1=10e-
(C) F-= 9+1 = 10
O2-= 8+2 = 10
N3-= 7+3 = 10
Mg2+=12-2=10
(d) Li+=3-1=2e-
Na+=11-1=10e-
O-2=8+2=10e-
F-=9+1=10e-
Hence, N3-, O2-, F-, Na+ are isoelectronic.

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2019 (10 Apr Shift 1)
The isomeric deuterated bromide with molecular formula C4H8DBr having two chiral carbon atoms is

Options:
A 2-Bromo-1-deuterobutane
B 2-Bromo-2-deuterobutane
C 2-Bromo-3-deuterobutane
D 2-Bromo-1-deutero-2-methylpropane

The correct answer is:
2-Bromo-3-deuterobutane

Solution:
The isomeric deuterated bromide with molecular formula C4H8DBr having two chiral carbon atoms is

2-Bromo-3-deuterobutane

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2023 (25 Jan Shift 2)
The isomer(s) of CoNH34Cl2] that has/ have a Cl-Co-Cl angle of 90o, is/are:

Options:
A meridional and trans
B cis and trans
C trans only
D cis only

The correct answer is:
cis only

Solution:
The isomer(s) of CoNH34Cl2] that has/ have a Cl-Co-Cl angle of 90o, is the cis-isomer only. The structure of the cis isomers is written below:

Here, as we can see that one 90o angle of Cl-Co-Cl bond is present.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2020 (09 Jan Shift 2)
The isotopes of hydrogen are:

Options:
A Protium, deuterium and tritium
B Tritium and protium only
C Protium and deuterium only
D Deuterium and tritium only

The correct answer is:
Protium, deuterium and tritium

Solution:
Isotopes have same atomic number and different atomic mass i.e., Same number of proton different number of Neutron relative atomic mass
Protium H11
Deuterium H21
Tritium H31

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2019 (09 Jan Shift 1)
The isotopes of hydrogen have different physical properties due to difference in mass. They have almost same chemical properties with a difference in their rates of reactions which is mainly due to

Options:
A their different enthalpy of bond dissociation
B different physical properties
C different atomic masses
D different electronic configurations

The correct answer is:
their different enthalpy of bond dissociation

Solution:
Isotopes have same chemical properties, but heavier isotope reacts with slower reaction rate. For example, Deuterium reaction slower than hydrogen.

Topic: Chemistry, s Block Elements, NEET
The IUPAC name for the complex $\left[\mathrm{Co}\left(\mathrm{NO}_2\right)\left(\mathrm{NH}_3\right)_5\right] \mathrm{Cl}_2$ is

Options:
A
nitrito-N-pentaamminecobalt (III) chloride

B
nitrito-N-pentaamminecobalt (II) chloride

C
pentaammine nitrito- $\mathrm{N}$-cobalt (II) chloride

D
pentaammine nitrito- $\mathrm{N}$-cobalt (III) chloride

The correct answer is:
pentaammine nitrito- $\mathrm{N}$-cobalt (III) chloride

Solution:
No Solution Available

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2006
The IUPAC name for the following compound is:

Options:
A 3-methyl-4-(1-methylprop-2-ynyl)-1-heptene
B 3, 5-dimethyl-4-propylhept-6-en-1-yne
C 3, 5-dimethyl-4-propyl hept-1-en-6-yne
D 3-methyl-4-(3-methylprop-1-enhyl)-1-heptyne

The correct answer is:
3, 5-dimethyl-4-propyl hept-1-en-6-yne

Solution:
3, 5-dimethyl-4-propyl hept-1-en-6-yne

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2019 (12 Apr Shift 2)
The IUPAC name for the following compound is :

Options:
A 2,5-dimenthyl-5-carboxy-hex-3-enal
B 2,5-dimenthyl-6-carboxy-hex-3-enal
C 2,5-dimenthyl-6-carboxy-hex-3-enoic acid
D 6-formyl-2-methy 1-hex-3-enoic acid

The correct answer is:
2,5-dimenthyl-6-carboxy-hex-3-enoic acid

Solution:
2. 5-dimenthyl-6-carboxy-hex-3-enoic acid

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2020 (02 Sep Shift 1)
The IUPAC name for the following compound is

Options:
A 2 -propylhex-1-ene
B 2 -butylpent-1-ene
C 2-propyl-2- butylethene
D Propy1-1-butylethene

The correct answer is:
2 -propylhex-1-ene

Solution:
2 propyl hex $-1$ ene

Topic: Chemistry, General Organic Chemistry, JEE Main
The IUPAC name for the hydrocarbon represented by the Swastika sign is

Options:
A Neononane
B Tetraethylcarbon
C 2-ethylpentane
D \( 3,3 \) - diethylpentane

The correct answer is:
\( 3,3 \) - diethylpentane

Solution:
A bond line structure is defined as the representation of the structure of organic compound though lines depicting the bonds. When the condensed structure is drawn based on the Swastik symbol it gives the structure of an organic compound called 3,3-diethyl pentane as given in the attachment. The common name of the compound with the Swastika symbol is tetraethylmethane.

Topic: Chemistry, General Organic Chemistry, JEE Main
The IUPAC name of $\mathrm{CH}_3 \mathrm{COCH}\left(\mathrm{CH}_3\right)_2$ is

Options:
A
2-methyl-3-butanone

B
4-methylisopropyl ketone

C
3-methyl-2-butanone

D
Isopropylmethyl ketone

The correct answer is:
3-methyl-2-butanone

Solution:
No Solution Available

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2003
The IUPAC name of ethylidene chloride is

Options:
A 1-chloroethene
B 1, 2-dichloroethane
C 1, 1-dichloroethane
D 1-chloroethyne

The correct answer is:
1, 1-dichloroethane

Solution:
1,1-dichloroethane

Ethylidene chloride is a geminal dihalide with two carbon atoms. In common name system, gem-dihalides are named as alkylidene halides and vic-dihalides are named as alkylene dihalides. The structure of ethylidene chloride is:

CH3−CHCl2.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2022 (25 Jun Shift 1)
The IUPAC name of following compound is :

Options:
A 2-nitro-4-hydroxymethyl-5-amino benzaldehyde
B 3-amino-4-hydroxymethyl-5-nitrogenzaldehyde
C 5-amino-4-hydroxymethyl-2-nitrobenzaldehyde
D 4-amino-2-formyl-5-hydroxymethyl nitrogenzene

The correct answer is:
5-amino-4-hydroxymethyl-2-nitrobenzaldehyde

Solution:
5-Amino-4-(hydroxymethyl)-2-nitro benzene carbaldehyde.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2020 (06 Sep Shift 2)
The IUPAC name of
is

Options:
A
$1, 1-$diethyl$-2,2-$dimethylpentane

B
$4, 4-$dimethyl$-5, 5-$diethylpentane

C
$5, 5-$diethyl$-4, 4-$diemthylpentane

D
$3-$ethyl$-4, 4-$dimethylheptane

The correct answer is:
$3-$ethyl$-4, 4-$dimethylheptane

Solution:
The correct answer is $3-$ethyl$-4, 4-$dimethylheptane.
Hence, (D) is correct.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2007
The IUPAC name of K3[Co(C2O4)3] is:

Options:
A Potassium tris(oxalato)cobaltateIII
B Potassium tris(oxalato)cobalt(III)
C Potassium trioxalatocobalt(III)
D Potassium trioxalatocobaltateIII

The correct answer is:
Potassium trioxalatocobaltateIII

Solution:
Ligands that include a numerical prefix in the name use the prefixes bis for 2, tris for 3, or tetrakis for 4 to indicate their number.

If the complex ion is an anion, we drop the ending of the metal name and add –ate.

In the given coordination compound, Co is the cobalt metal, C2O42- are the ligands named as oxalato.

The charge on [Co(C2O4)3] is -3 and oxalate is having charge of -2.

X+3(-2)=-3X=+3

Hence, the IUPAC name of the compound K3[Co(C2O4)3] is Potassium trioxalatocobaltate(III).

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2023 (06 Apr Shift 2)
The I.U.P.A.C. name of $\mathrm{K}_3\left[\operatorname{Ir}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]$ is

Options:
A Potassium tri oxalato iridium (III)
B Potassium tri oxalato iridate (III)
C Potassium tris (oxalato) iridium (III)
D Potassium tris (oxalato) iridate (III)

The correct answer is:
Potassium tri oxalato iridate (III)

Solution:
The complex $\mathrm{K}_3\left[\operatorname{Ir}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]$ consists of an iridium center surrounded by three oxalate $\left(\mathrm{C}_2 \mathrm{O}_4\right)$ ligands. The iridium is in the +3 oxidation state. The complex ion is an anion, thus the name ends with "iridate". The correct IUPAC name is Potassium tri oxalato iridate (III).

Topic: Chemistry, Coordination Compounds, JEE Main
The IUPAC name of neopentane is

Options:
A
2-methylbutane

B
2, 2-dimethylpropane

C
2-methylpropane

D
2,2-dimethylbutane

The correct answer is:
2, 2-dimethylpropane

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2009
The IUPAC name of $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ is

Options:
A Tetra carbonyl nickel (II)
B Tetra carbonyl nickel (0)
C Tetra carbonyl nickelate (II)
D Tetra carbonyl nickelate (0)

The correct answer is:
Tetra carbonyl nickel (0)

Solution:
The complex $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ consists of a nickel atom coordinated to four carbon monoxide $\mathrm{CO}$ ligands. In this complex, nickel is in the zero oxidation state, as carbon monoxide is a neutral ligand. Thus, the correct IUPAC name is Tetra carbonyl nickel (0).

Topic: Chemistry, Coordination Compounds, JEE Main
The IUPAC name of the complex \( \left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Br}\left(\mathrm{NO}_{2}\right) \mathrm{Cl}\right] \mathrm{Cl} \) is

Options:
A Triamminechlorobromonitroplatinum (IV) chloride.
B Triamminebromonitrochloroplatinum (IV) chloride
C Triamminebromidochloridonitroplatinum (IV) chloride
D Triamminenitrochlorobromoplatinum (IV) chloride

The correct answer is:
Triamminebromidochloridonitroplatinum (IV) chloride

Solution:
Ligand	Name Or (Alphabetical Order)

NH3	ammine 1

Br-	bromide/Bromo 2

Cl-	Chloride/Chloro 3

NO2-

Nitrito-N 4

So, the name of the complex is triamminebromidochloridoplatinum (iv) chloride.

Topic: Chemistry, Coordination Compounds, JEE Main
The IUPAC name of the complex PtNH32ClNH2CH3Cl is

Options:
A Diamminechlorido (methanamine) platinum II chloride
B Diammine(methanamine)chlorido platinum II chloride
C Diamminechlorido (aminomethane) platinum II chloride
D Bisammine (methanamine) chlorido platinum II chloride

The correct answer is:
Diamminechlorido (methanamine) platinum II chloride

Solution:
NH 3 is neutral ligand, its IUPAC name is ammine
Cl is negative ligand, its IUPAC name is chlrido
NH 2 CH 3 is neutral ligand, its IUPAC name is methanamine
Oxidation state of platinum ⇒ x+2×0+1 −1 +1 0 =1

x=2
Naming of ligand should be done in the alphabetically order.

The IUPAC name of the complex is Diamminechlorido (methanamine) platinum II chloride

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2020 (07 Jan Shift 1)
The IUPAC name of the complex [Pt(NH3)2ClNH2CH3]Cl is

Options:
A Diamminechlorido(aminomethane)platinum(II) chloride
B Diammine(methanamine)chloridoplatinum(II) chloride
C Diamminechlorido(methanamine)platinum(II) chloride
D Bisammine(methanamine)chloridoplatinum(II) chloride

The correct answer is:
Diamminechlorido(methanamine)platinum(II) chloride

Solution:
PtNH32ClNH2CH3Cl
Let x be the oxidation number of Pt
x+0â1+0=+1
x=+2â Pt+2
Diamminechlorido(methanamine)platinum(II)chloride

Topic: Chemistry, Coordination Compounds, JEE Main
The IUPAC name of the complex $\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]$ is

Options:
A Platinum (II) diammino dichloride
B Diammino dichloro platinate (IV)
C Bis (ammino) dichloro platinum (IV)
D Dichloro diammine platinum (II)

The correct answer is:
Dichloro diammine platinum (II)

Solution:
In the complex $\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]$, platinum is in the +2 oxidation state. The ammonia $\left(\mathrm{NH}_3\right)$ ligands are named as "ammine" and the chloride ($\mathrm{Cl}$) ligands are named as "chloro". According to IUPAC naming rules, the ligands are listed alphabetically before the central metal ion, and the oxidation state of the metal is indicated in Roman numerals. Therefore, the correct name is Dichloro diammine platinum (II).

Topic: Chemistry, Coordination Compounds, JEE Main
The IUPAC name of the compound

Options:
A 1-bromo-1-chloro-4, 4 '-bicyclobutane
B 4 -(4'-chlorocyclobutyl)-1-bromocyclobutane
C 3 -bromo- 3 '-chloro-1, 1 '-bicyclobutane
D 4-(4'-bromocyclobutyl)-1-chlorocyclo butane

The correct answer is:
3 -bromo- 3 '-chloro-1, 1 '-bicyclobutane

Solution:
Br gets preference over Cl. Hence, its IUPAC name is 3 -bromo-3'-chloro-1, 1 '-bicyclobutane.

Topic: Chemistry, General Organic Chemistry, JEE Main
The IUPAC name of the compound
< smiles>CC1CO1 < /smiles>
is

Options:
A
1, 2-Propoxide

B
Propylene oxide

C
1,2-Oxo propane

D
1, 2-Epoxy propane

The correct answer is:
1, 2-Epoxy propane

Solution:
1, 2-Epoxy propane is the correct IUPAC name of given compound.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2012 (07 May Online)
The IUPAC name of the compound CH2 = CH — CH(CH3​ )2 is

Options:
A 1,1 -dimethyl 1-2-butene
B 3-methyl-1-butene
C 2-vinyl propane
D None of these

The correct answer is:
3-methyl-1-butene

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
The IUPAC name of the compound $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHC} \equiv \mathrm{CH}$ is

Options:
A pent-4-yn-2-ene
B pent-3-en-1-yne
C pent-2-en-4-yne
D pent-1-yn-3-ene

The correct answer is:
pent-3-en-1-yne

Solution:
$$
\stackrel{5}{\mathrm{C}_3}-\stackrel{4}{\mathrm{C}} \mathrm{H}=\stackrel{3}{\mathrm{C}} \mathrm{H}-\stackrel{2}{\mathrm{C}} \equiv \stackrel{1}{\mathrm{C}} \mathrm{H}
$$
pent-3-en-1-yne

Topic: Chemistry, General Organic Chemistry, NEET, NEET 2010 (Mains)
The IUPAC name of the compound $\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHC} \equiv \mathrm{CH}$ is

Options:
A pent-4-yn-2-ene
B pent-3-en-1-yne
C pent-2-en-4-yne
D pent-1-yn-3-ene

The correct answer is:
pent-3-en-1-yne

Solution:
$\stackrel{5}{\mathrm{C}} \mathrm{H}_{3}-\stackrel{4}{\mathrm{C}} \mathrm{C}=\stackrel{3}{\mathrm{C}} \mathrm{H}-\stackrel{2}{\mathrm{C}}=\stackrel{1}{\mathrm{C}} \mathrm{H}$

Topic: Chemistry, General Organic Chemistry, JEE Main
The IUPAC name of the compound having the formula $\mathrm{CH} \equiv \mathrm{C}-\mathrm{CH}=\mathrm{CH}_2$ is -

Options:
A 1-butene-3-yne
B 3-buten-1-yne
C 1-butyn-3-ene
D but-1-yn-3-ene

The correct answer is:
1-butene-3-yne

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
The IUPAC name of the compound having the formula $\mathrm{CH} \equiv \mathrm{C}-\mathrm{CH}=\mathrm{CH}_2$ is -

Options:
A 1-butene-3-yne
B 3-buten-1-yne
C 1-butyn-3-ene
D but-1-yn-3-ene

The correct answer is:
1-butene-3-yne

Solution:

Topic: Chemistry, General Organic Chemistry, NEET, NEET 2009 (Mains)
The IUPAC name of the compound is:

Options:
A 1-amino-1-phenyl-2-methylpropane
B 2-methyl-1-phenylpropan-1-amine
C 2-methyl-1-amino-1-phenylpropane
D 1-isopropyl-1-phenylmethyl amine

The correct answer is:
2-methyl-1-phenylpropan-1-amine

Solution:
a main chain of three carbon with amine as principle group and phenyl and methyl as substituents. Thus, it is-
2-methyl-1-phenylpropan-1-amine

Topic: Chemistry, Amines, NEET
The IUPAC name of the compound NH42NiC2O42H2O2 is

Options:
A Nickel (II) diammino dioxalato diaquate
B Dioxalato diammino diaquo nickelate (III)
C Ammonium diaquabis (oxalato) nickelate (II)
D Ni dioxalato diaqua (II) amminate

The correct answer is:
Ammonium diaquabis (oxalato) nickelate (II)

Solution:
The set of rules for naming a coordination compound is:

1. Name the ligands, put in alphabetical order.

2. Add prefixes for ligands that appear more than once.

3. Name the metal

For Cationic complex- write metal name.

For anionic complex- write metal prefix +ate.

4. Add oxidation state in Roman numerals.

5. As separate word, write "ion", if not neutral, or list counterions (positive first, negative second).

The IUPAC name of the compound NH42NiC2O42H2O2 is Ammonium diaquabis (oxalato) nickelate (II).

Topic: Chemistry, d and f Block Elements, JEE Main
The IUPAC name of the compound

Options:
A
3, 3- dimethyl -1- hydroxy cyclohexane

B
1,1-dimethyl -3- cyclohexanol

C
3,3- dimethyl -1- cyclohexanol

D
1,1 - dimethyl -3- hydroxy cyclohexane

The correct answer is:
3,3- dimethyl -1- cyclohexanol

Solution:
No Solution Available

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2004
The IUPAC name of the coordination compound $\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]$ is

Options:
A
Potassium hexacyanoferrate (II)

B
Potassium hexacyanoferrate (III)

C
Potassium hexacyanoiron (II)

D
tripotassium hexcyanoiron (II)

The correct answer is:
Potassium hexacyanoferrate (III)

Solution:
No Solution Available

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2005
The IUPAC name of the following compound is

Options:
A
(E)-2-hepten-4-yne

B
(Z)-5-hepten-3-yne

C
(E)-5-hepten-3-yne

D
(Z)-2-hepten-4-yne

The correct answer is:
(E)-2-hepten-4-yne

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2012 (19 May Online)