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

Did you get this right?

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.

Your note

JEE Main Inorganic Chemistry in other years

Inorganic Chemistry 2026 Inorganic Chemistry 2025 Inorganic Chemistry 2024 Inorganic Chemistry 2023 Inorganic Chemistry 2022 Inorganic Chemistry 2021 Inorganic Chemistry 2020 Inorganic Chemistry 2018 Inorganic Chemistry 2017 Inorganic Chemistry 2016 Inorganic Chemistry 2015 Inorganic Chemistry 2014

Keep practicing Inorganic Chemistry

Work through every JEE Main Inorganic Chemistry PYQ, year by year.

Browse all Chemistry topics

JEE Main 2019 — Chemistry Inorganic Chemistry

2019mcqmedium

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

Did you get this right?

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.

JEE Main 2019 — Chemistry Inorganic Chemistry

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

JEE Main 2019 — Chemistry Inorganic Chemistry

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