1

JEE Advanced 2021 Paper 1 Online

MCQ (Single Correct Answer)
The calculated spin only magnetic moments of [ Cr(NH3)6]3+ and [CuF6]3$$-$$ in BM, respectively, are

(Atomic numbers of Cr and Cu are 24 and 29, respectively)
A
3.87 and 2.84
B
4.90 and 1.73
C
3.87 and 1.73
D
4.90 and 2.84

Explanation

$${[Cr{(N{H_3})_6}]^{3 + }}$$

Atomic number of Cr = 24

Electron configuration of Cr = 1s2 2s2 2p6 3s2 3p6 3d5 4s1

Electronic configuration of Cr3+ = 1s2 2s2 2p6 3s2 3p6 3d3 4s0

NH3 is a weak field ligand. The splitting of d-orbital occur.


Number of unpaired electrons, n = 3

Magnetic moment (spin only), $${\mu _s} = \sqrt {n(n + 2)} BM$$

$$ = \sqrt {3(3 + 2)} = \sqrt {3 \times 5} = \sqrt {15} = 3.87BM$$

$${[Cu{F_6}]^{3 - }}$$ Atomic number of Cu = 29

Electronic configuration of Cu = 1s2 2s2 2p6 3s2 3p6 3d10 4s1

Electronic configuration of Cu2+ = 1s2 2s2 2p6 3s2 3p6 3d8 4s0

F$$-$$ is a weak field ligand, so splitting of d-orbitals occur as follows:


Number of unpaired electrons, n = 2

Spin only magnetic moment, $${\mu _s} = \sqrt {n(n + 2)} BM$$

$$ = \sqrt {2(2 + 2)} = \sqrt {2 \times 4} = 2\sqrt 2 = 2.84BM$$
2

JEE Advanced 2018 Paper 2 Offline

MCQ (Single Correct Answer)
Match each set of hybrid orbitals from LIST - A with complex(es) given in LIST - B

List - A List - B
P. dsp2 1. [FeF6]4-
Q. sp3 2. [Ti(H2O)3Cl3]
R. sp3d2 3. [Cr(NH3)6]3+
S. d2sp3 4. [FeCl4]2-
5. Ni(CO)4
6. [Ni(CN)4]2-


The correct option is
A
$$P - 5;Q - 4,6;R - 2,3;S - 1$$
B
$$P - 5,6;Q - 4;R - 3;S - 1,2$$
C
$$P - 6;Q - 4,5;R - 1;S - 2,3$$
D
$$P - 4,6;Q - 5,6;R - 1,2;S - 3$$
3

JEE Advanced 2016 Paper 1 Offline

MCQ (Single Correct Answer)

Among [Ni(CO)4], [NiCl4]2$$-$$, [Co(NH3)4)Cl2], Na3[CoF6], Na2O2 and CsO2, the total number of paramagnetic compound is

A
2
B
3
C
4
D
5
4

JEE Advanced 2014 Paper 2 Offline

MCQ (Single Correct Answer)

Match each coordination compound in List I with an appropriate pair of characteristics from List II and select the correct answer using the code given below the lists.

{en = H2NCH2CH2NH2; atomic numbers : Ti = 22, Cr = 24; Co = 27; Pt = 78}

List I List II
P. $$[Cr{(N{H_3})_3}C{l_2}]Cl$$
1. Paramagnetic and exhibits ionisation isomerism.
Q. $$[Ti{({H_2}O)_5}Cl]{(N{O_3})_2}$$
2. Diamagnetic and exhibits cis-trans isomerism.
R. $$[Pt(en)(N{H_3})Cl]N{O_3}$$
3. Paramagnetic and exhibits cis-trans isomerism.
S. $$[Co{(N{H_3})_4}{(N{O_3})_2}]N{O_3}$$
4. Diamagnetic and exhibits ionisation isomerism.

A
P-4, Q-2, R-3, S-1
B
P-3, Q-1, R-4, S-2
C
P-2, Q-1, R-3, S-4
D
P-1, Q-3, R-4, S-2

Explanation

For [Cr(NH3)4Cl2]Cl,

For [Ti(H2O)5Cl](NO3)2,

The complex is paramagnetic due to presence of unpaired electron.

Since the groups between the complex ion and ions outside can be exchanged, it shows ionisation isomerism.

For [Pt(en)(NH3)Cl]NO3,

The complex is square planar and all electrons are paired. So it is diamagnetic. Since the groups between the complex ion and ions outside can be exchanged, it shows ionisation isomerism.

For [Co(NH3)4(NO3)2]NO3, NH3 is a strong ligand and causes pairing

Joint Entrance Examination

JEE Main JEE Advanced WB JEE

Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

Medical

NEET

CBSE

Class 12