JEE MAIN - Chemistry (2021 - 24th February Evening Shift - No. 8)
The calculated magnetic moments (spin only value) for species $${[FeC{l_4}]^{2 - }}$$, $${[Co{({C_2}{O_4})_3}]^{3 - }}$$ and $$MnO_4^{2 - }$$ respectively are :
5.92, 4.90 and 0 BM
4.90, 0 and 1.73 BM
5.82, 0 and 0 BM
4.90, 0 and 2.83 BM
Explanation
(i) $${[FeC{l_4}]^{2 - }}\buildrel {} \over
\longrightarrow F{e^{2 + }}\buildrel {} \over
\longrightarrow [Ar]3{d^6}$$
$$\therefore$$ Cl is weak field ligand so does not pairing occur
_24th_February_Evening_Shift_en_8_1.png)
So, magnetic moment $$(\mu ) = \sqrt {n(n + 2)} BM$$
$$ = \sqrt {4(4 + 2)} BM$$ (n = Number of total unpaired e$$-$$ = 4)
$$ = \sqrt {24} BM = 4.90\,BM$$
(ii) $${[Co{({C_2}{O_4})_3}]^{3 - }}\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} C{o^{3 + }}\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} [Ar]3{d^6}$$
C2O4 is strong field ligand so pairing occur.
_24th_February_Evening_Shift_en_8_2.png)
All electrons are paired, n = 0
hence, $$\mu$$ = 0
(iii) $$MnO_4^{2 - }\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} M{n^{ + 6}}\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} [Ar]3{d^1}$$
$$n = 1$$
$$ = \sqrt {n(n + 2)} BM$$
$$ = \sqrt {1(1 + 2)} BM$$
$$ = \sqrt 3 BM$$
$$ = 1.73\,BM$$
$$\therefore$$ Cl is weak field ligand so does not pairing occur
So, magnetic moment $$(\mu ) = \sqrt {n(n + 2)} BM$$
$$ = \sqrt {4(4 + 2)} BM$$ (n = Number of total unpaired e$$-$$ = 4)
$$ = \sqrt {24} BM = 4.90\,BM$$
(ii) $${[Co{({C_2}{O_4})_3}]^{3 - }}\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} C{o^{3 + }}\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} [Ar]3{d^6}$$
C2O4 is strong field ligand so pairing occur.
All electrons are paired, n = 0
hence, $$\mu$$ = 0
(iii) $$MnO_4^{2 - }\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} M{n^{ + 6}}\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} [Ar]3{d^1}$$
$$n = 1$$
$$ = \sqrt {n(n + 2)} BM$$
$$ = \sqrt {1(1 + 2)} BM$$
$$ = \sqrt 3 BM$$
$$ = 1.73\,BM$$
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