The correct decreasing order of spin only magnetic moment values (BM) of Cu+, Cu2+, Cr2+ and Cr3+ ions is :
Solution
<p>$\mathrm{Cu}^{+}:[\mathrm{Ar}] 3 \mathrm{~d}^{10}$, Spin only magnetic moment $=0$ B.M.</p>
<p>$\mathrm{Cu}^{+2}:[\mathrm{Ar}] 3 \mathrm{~d}^9$, Spin only magnetic moment $=\sqrt{3}$ B.M.</p>
<p>$\mathrm{Cr}^{+2}:[\mathrm{Ar}] 3 \mathrm{~d}^4$, Spin only magnetic moment $=\sqrt{24}$ B.M.</p>
<p>$\mathrm{Cr}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^3$, Spin only magnetic moment $=\sqrt{15}$ B.M.</p>
<p>Order of $\mu: \mathrm{Cr}^{+2}>\mathrm{Cr}^{+3}>\mathrm{Cu}^{+2}>\mathrm{Cu}^{+}$</p>
About this question
Subject: Chemistry · Chapter: d and f Block Elements · Topic: Properties of Transition Metals
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