The correct order of the calculated spin-only magnetic moments of complexes (A) to (D) is: (A) Ni ( CO ) _ 4 (B) [Ni ( H _ 2 O) _ 6 ]C l _ 2 (C) N a _ 2 [Ni (CN) _ 4 ] (D) PdC l _ 2 (PP h _ 3 ) _ 2

Correct answer: D. (A) (C) (D) < (B) [Pd (PP h _ 3 ) _ 2 C l _ 2 ], P d ^ 2 + = 4 d ^ 8 , M = 0 [Ni (CO) _ 4 ], Ni = 3 d ^ 8 4 s ^ 2 (SFL), M = 0 [Ni (CN) _ 4 ] ^ 2 - N i ^ 2 + = 3 d ^ 8 (SFL) M = 0 [Ni ( H _ 2 O) _ 6 ] ^ 2 + N i ^ 2 + = 3 d ^ 8 (WFL) t _ 2g ^ 2, 2, 2 , e _ g ^ 1, 1 So, unpaired electron is 2, M = 8 BM

JEE Main 20208 JanuaryEvening Shift

Exam: JEE Main
Year: 2020
Date: 8 January
Shift: Evening Shift
Subject: Chemistry
Chapter: Aldehydes and Ketones

JEE Main 2020 · 8 January · Evening Shift · Chemistry · Aldehydes and Ketones

The correct order of the calculated spin-only magnetic moments of...

The correct order of the calculated spin-only magnetic moments of complexes

\left(A\right)

to \left(D\right)

is:<br />\left(A\right) Ni{\left(\text{ CO }\right)}

_{4}

<br />\left(B\right)

\left[Ni{\left({H}_{2}O\right)}_{6}\right]C{l}_{2}

<br />\left(C\right) N{a}_{2}\left[Ni{\left(CN\right)}_{4}\right]

\left(D\right) PdC{l}_{2}{\left(PP{h}_{3}\right)}_{2}

Options

A $\left(A\right) \approx \left(C\right) < \left(B\right) \approx \left(D\right)$
B $\left(C\right) < \left(D\right) < \left(B\right) < \left(A\right)$
C $\left(C\right) \approx \left(D\right) < \left(B\right) < \left(A\right)$
D $\left(A\right) \approx \left(C\right) \approx \left(D\right) < \left(B\right)$

Correct answer

D. $\left(A\right) \approx \left(C\right) \approx \left(D\right) < \left(B\right)$

Step-by-step solution

\left[Pd{\left(PP{h}_{3}\right)}_{2}C{l}_{2}\right], P{d}^{2 +} = 4{d}^{8}, M = 0

\left[Ni{\left(CO\right)}_{4}\right], Ni = 3{d}^{8}4{s}^{2}\left(SFL\right), M = 0

<br/>

{\left[Ni{\left(CN\right)}_{4}\right]}^{2 -} N{i}^{2 +} = 3{d}^{8}\left(SFL\right) M = 0

<br/>

{\left[Ni{\left({H}_{2}O\right)}_{6}\right]}^{2 +} N{i}^{2 +} = 3{d}^{8}\left(WFL\right) {t}_{2g}^{2, 2, 2}, {e}_{g}^{1, 1}

So, unpaired electron is

2, M = \sqrt{8}BM

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