The correct order of the spin… - JEE Main 2019 Chemistry PYQ

JEE Main 2019 — Chemistry Inorganic Chemistry

2019mcqmedium

The correct order of the spin –only magnetic moment of metal ions in the following low-spin complexes, ${[V{(CN)}_{6}]}^{4-},{[Fe{(CN)}_{6}]}^{4-},{[Ru{(N{H}_{3})}_{6}]}^{3+}$ and ${[Cr{(N{H}_{3})}_{6}]}^{2+},$ is:

Official previous-year question

Held on 8 Apr 2019 · Verified 6 Jul 2026.

Options

A
${ V}^{2}+>C{r}^{2+}>R{u}^{3+}>F{e}^{2+}$
B
$C{r}^{2+}>{V}^{2}+>R{u}^{3+}>F{e}^{2+}$
C
$C{r}^{2+}>R{u}^{3+}>F{e}^{2+}>{V}^{2+}$
D
${V}^{2+}>R{u}^{3+}>C{r}^{2+}>F{e}^{2+}$

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Solution

(i) ${V}^{+2}(23)\Longrightarrow [Ar]3{d}^{3}4{d}^{0}\Longrightarrow$ JEE Main 2019 Chemistry, Coordination Compounds — solution figure 1 $(n=3)$ $\Rightarrow \mu \sqrt{3(3+2)}=3.87$

(ii) $C{r}^{+2}(24)\Longrightarrow [Ar]3{d}^{4}$ JEE Main 2019 Chemistry, Coordination Compounds — solution figure 2 $(n=2)$ $\Rightarrow \mu \sqrt{2(2+2)}=2.82$

(iii) $R{u}^{+3}(44)\Longrightarrow [kr]4{d}^{5}$ JEE Main 2019 Chemistry, Coordination Compounds — solution figure 3 $(n=1)$ $\Rightarrow \mu \sqrt{1(1+2)}=3$

(iv) $F{e}^{+2}(26) \Longrightarrow [Ar]3{d}^{6}$ JEE Main 2019 Chemistry, Coordination Compounds — solution figure 4 $(n=0)$ $\Rightarrow \mu =0$

Order of magnetic moment will be $\propto$ no. of unpaired ${e}^{-}$

${V}^{+2}> C{r}^{+2}>R{u}^{+3} >F{e}^{+2}$

$\mu =\sqrt{n(n+2)}$ B.M. Hence more the value of $n$ more will be magnitude moment.

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Options

${ V}^{2}+>C{r}^{2+}>R{u}^{3+}>F{e}^{2+}$

$C{r}^{2+}>{V}^{2}+>R{u}^{3+}>F{e}^{2+}$

$C{r}^{2+}>R{u}^{3+}>F{e}^{2+}>{V}^{2+}$

${V}^{2+}>R{u}^{3+}>C{r}^{2+}>F{e}^{2+}$

Solution

(i) ${V}^{+2}(23)\Longrightarrow [Ar]3{d}^{3}4{d}^{0}\Longrightarrow$ JEE Main 2019 Chemistry, Coordination Compounds — solution figure 1 $(n=3)$ $\Rightarrow \mu \sqrt{3(3+2)}=3.87$

(ii) $C{r}^{+2}(24)\Longrightarrow [Ar]3{d}^{4}$ JEE Main 2019 Chemistry, Coordination Compounds — solution figure 2 $(n=2)$ $\Rightarrow \mu \sqrt{2(2+2)}=2.82$

(iii) $R{u}^{+3}(44)\Longrightarrow [kr]4{d}^{5}$ JEE Main 2019 Chemistry, Coordination Compounds — solution figure 3 $(n=1)$ $\Rightarrow \mu \sqrt{1(1+2)}=3$

(iv) $F{e}^{+2}(26) \Longrightarrow [Ar]3{d}^{6}$ JEE Main 2019 Chemistry, Coordination Compounds — solution figure 4 $(n=0)$ $\Rightarrow \mu =0$

Order of magnetic moment will be $\propto$ no. of unpaired ${e}^{-}$

${V}^{+2}> C{r}^{+2}>R{u}^{+3} >F{e}^{+2}$

$\mu =\sqrt{n(n+2)}$ B.M. Hence more the value of $n$ more will be magnitude moment.

Options

Did you get this right?

Solution

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JEE Main Inorganic Chemistry in other years

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Physical Chemistry

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Options

Did you get this right?

Solution

Your note

JEE Main Inorganic Chemistry in other years

More JEE Main Inorganic Chemistry questions

Other Chemistry topics for JEE Main

Physical Chemistry

Organic Chemistry

Keep practicing Inorganic Chemistry