Chemistry - GENERAL CHEMISTRY Question with Solution | TestHub

To find the elements, we use the given information. Element X is in the 4th period, 8th group, which is Iron (Fe), so A = 26.

Given $A-B=2$, so $26-B=2⟹B=24$. Element Y is Chromium (Cr).

Given $\frac{A+B}{2}=C-2$, so $\frac{26+24}{2}=C-2⟹25=C-2⟹C=27$. Element Z is Cobalt (Co).

Now, we determine the number of unpaired electrons and calculate the magnetic moment ($\mu =\sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons).

Element X (Fe): [Ar] $3d^{6}4s^2$.

Fe ($n=4$): $\mu =\sqrt{4(4+2)}=\sqrt{24}\approx 4.90$ BM

Fe²⁺ ($n=4$): $\mu =\sqrt{4(4+2)}=\sqrt{24}\approx 4.90$ BM

Fe³⁺ ($n=5$): $\mu =\sqrt{5(5+2)}=\sqrt{35}\approx 5.92$ BM

Element Y (Cr): [Ar] $3d^{5}4s^1$.

Cr ($n=6$): $\mu =\sqrt{6(6+2)}=\sqrt{48}\approx 6.93$ BM

Cr⁺ ($n=5$): $\mu =\sqrt{5(5+2)}=\sqrt{35}\approx 5.92$ BM

Cr²⁺ ($n=4$): $\mu =\sqrt{4(4+2)}=\sqrt{24}\approx 4.90$ BM

Cr³⁺ ($n=3$): $\mu =\sqrt{3(3+2)}=\sqrt{15}\approx 3.87$ BM

Element Z (Co): [Ar] $3d^{7}4s^2$.

Co ($n=3$): $\mu =\sqrt{3(3+2)}=\sqrt{15}\approx 3.87$ BM

Co²⁺ ($n=3$): $\mu =\sqrt{3(3+2)}=\sqrt{15}\approx 3.87$ BM

Co³⁺ ($n=4$): $\mu =\sqrt{4(4+2)}=\sqrt{24}\approx 4.90$ BM

Let's evaluate the options:

A. Y > X > Z²⁺

Cr ($n=6$, $\mu \approx 6.93$) > Fe ($n=4$, $\mu \approx 4.90$) > Co²⁺ ($n=3$, $\mu \approx 3.87$). This order is CORRECT.

B. Y⁺ > X²⁺ > Z²⁺

Cr⁺ ($n=5$, $\mu \approx 5.92$) > Fe²⁺ ($n=4$, $\mu \approx 4.90$) > Co²⁺ ($n=3$, $\mu \approx 3.87$). This order is CORRECT.

C. Y = X³⁺ > Z³⁺

Cr ($n=6$, $\mu \approx 6.93$) $\ne$ Fe³⁺ ($n=5$, $\mu \approx 5.92$). This order is INCORRECT.

D. X³⁺ > Y²⁺ > Z

Fe³⁺ ($n=5$, $\mu \approx 5.92$) > Cr²⁺ ($n=4$, $\mu \approx 4.90$) > Co ($n=3$, $\mu \approx 3.87$). This order is CORRECT.

The correct options are A, B, and D.