En = Total energy, Kn = Kinetic Energy, Vn = Potential Energy, rn = Radius of nth orbit
Match the following
| Column I | Column II | ||
|---|---|---|---|
| (A) | $$\frac{V_n}{K_n} = ?$$ | (P) | 0 |
| (B) | If radius of $$n^{\text{th}}$$ orbit $$\propto E_n^x$$, $$x = ?$$ | (Q) | −1 |
| (C) | Angular momentum in lowest orbital | (R) | −2 |
| (D) | $$\frac{1}{r_n} \propto Z^y = ?$$ | (S) | 1 |
The coordination of $$\mathrm{Ni^2+}$$ is 4.
$$\mathrm{NiCl_2+KCN}$$ (excess) $$\to$$ A (cyano complex)
$$\mathrm{NiCl_2+ conc.~HCl}$$ (excess) $$\to$$ B (chloro complex)
The IUPAC names of $$\mathbf{A}$$ and $$\mathbf{B}$$ are
The edge length of unit cell of a metal having molecular weight $$75 \mathrm{~g} \mathrm{~mol}^{-1}$$ is 5 $$\mathop A\limits^o $$ which crystallizes in cubic lattice. If the density is $$2 \mathrm{~g} / \mathrm{cc}$$, find the radius of metal atom. $$\left(\mathrm{N}_{\mathrm{A}}=6 \times 10^{23}\right)$$. Give the answer in $$\mathrm{pm}$$.
$$ \mathrm{B}(\mathrm{OH})_3+\mathrm{NaOH} \quad \mathrm{NaBO}_2+\mathrm{Na}\left[\mathrm{~B}(\mathrm{OH})_4\right] $$$+\mathrm{H}_2 \mathrm{O}$
How can this reaction be made to proceed in forward direction?
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