Consider the four sulphates: $F e S O_{4}, M n S O_{4}, C o S O_{4}$ and $Z n S O_{4}$ . Four separate capsules of these materials have been made containing $0.1$ mole of the substance and are suspended by a thread. A strong magnet is brought close to each capsule. The order of sulphates according to the strength of the response towards the magnet is:

M n S O_{4} > F e S O_{4} > C o S O_{4} > Z n S O_{4}

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F e S O_{4} > M n S O_{4} > C o S O_{4} > Z n S O_{4}

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C o S O_{4} > Z n S O_{4} > F e S O_{4} > M n S O_{4}

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C o S O_{4} > F e S O_{4} > M n S O_{4} > Z n S O_{4}

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Solution

The correct option is A $M n S O_{4} > F e S O_{4} > C o S O_{4} > Z n S O_{4}$
If the sulphate has a greater magnetic moment then the deflection will be greater in the presence of a magnetic field.
Magnetic moment, $μ = \sqrt{n (n + 2)} BM$
where $n$ is the number of unpaired electrons.
All the metals are in $+ 2$ oxidation state in their respective sulphate.
Electronic configuration of the given metals are:
$F e^{2 +} : [A r] 3 d^{6}; n = 4$
$M n^{2 +} : [A r] 3 d^{5}; n = 5$
$C o^{2 +} : [A r] 3 d^{7}; n = 3$
$Z n^{2 +} : [A r] 3 d^{10}; n = 0$

Greater the number of unpaired electrons, greater will be the magnetic moment.
Hence the order is: $M n S O_{4} > F e S O_{4} > C o S O_{4} > Z n S O_{4}$

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