In Cu, all the d-electrons are paired (3d¹⁰ 4s¹). In Cr, all the d-electrons are unpaired (3d⁵ 4s¹). Hence, d-d electron re~ulsions in Cu are much greater than those in Cr. Therefore, Cu atom is larger in size than Cr. In Cu²⁺ (3d⁹), d-d electron repulsions decrease due to presence of one unpaired d-electron. Moreover, the electrons are attracted by 29 protons of the nucleus whereas in Cr²⁺, three unpaired electrons are still present but they are attracted by ony 24 protons of the nucleus. Thus, Cu2+is smaller in size than Cr²⁺.

Suppose the final volume after dilution is V₂
Then \({ M }_{ 1 }{ V }_{ 1 }={ M }_{ 2 }{ V }_{ 2 }\quad or\quad { V }_{ 2 }=\frac { { M }_{ 1 }{ V }_{ 1 } }{ { M }_{ 2 } } \)
Volume of water required to be added = \({ V }_{ 2 }-{ V }_{ 1 }=\frac { { M }_{ 1 }{ V }_{ 1 } }{ { M }_{ 2 } } -{ V }_{ 1 }=\left( \frac { { M }_{ 1 } }{ { M }_{ 2 } } -1 \right) { V }_{ 1 }=\left( \frac { { M }_{ 1 }-{ M }_{ 2 } }{ { M }_{ 2 } } \right) { V }_{ 1 }\)

Proteins have some structural similarity with synthetic polyamide.Both have \(-\overset{\overset{O}||}{C}-NH-\)peptide linkage

(i) HNO₂ (Nitrous acid)
(ii) HNO₃ (Nitric acid)
(iii) H₂N₂O₂ (Hydronitrous acid)
\(\underset{+3}{3HNO_2 } \xrightarrow{Disproportionation} \underset {+5}{HNO_3} + H_2O + \underset {+2}{2NO}\) 

\(\) 

(i) CL₂O
2x - 2 = 0
2x = 2
x = + 1
(ii) KBrO₃
+1 + x -6 = 0
x = +5
(iii) NaCIO₄
1 + x - 8 = 0
x = +7
(iv) CIO₂
x - 4 = 0
x = +4

\(k=\frac { 1 }{ R } \times \frac { l }{ a } \), where 'k' is conductivity of solution, 'R' is resistance, \(\frac { l }{ a } \) is cell constant. \({ \Lambda }_{ m }=\frac { 1000k }{ M } \), where 'k' is conductivity in S cm^-1, 'M' is molarity of solution. \({ \Lambda }_{ m }\) is molar conductivity in S cm² mol^-1.

Complex reaction proceeds through several elementary reactions.  Molecularity of each elementary reaction may be different, therefore, molecularity of complex reaction can't be determined.  Order of complex reaction is determined by slowest step in mechanism (involving elementary reactions).

Let the molecular mass of the compound be M_B.
\({p_A^0-p_A\over p_A^0}=X_B={{W_B\over M_B}\over {W_A\over M_A}+{W_B\over M_B}}\)
\(\Rightarrow{4.24-4.16\over 4.24}={{5\over M_b}\over{100\over 18}+{5\over M_B}}\)
\(\Rightarrow{0.08\times 5\over M_B}+0.08\times{100\over 18}={5\over M_B}\times 4.24\)
\(\Rightarrow{0.40\over M_B}+{8\over 18}={21.20\over M_B}\Rightarrow {20.80\over M_B}={8\over 18}\)
\(\Rightarrow\ M_B={20.80\times18\over 8}=46.8g\ mol^{-1}\)

In ccp structure, there are 8 tetrahedral voids. In dose-packed structure, there are eight spheres in the corner of the unit cell and each sphere is in contact with three others giving rise to eight tetrahedral voids.

\({ a }^{ 3 }=\frac { M\times Z }{ \rho \times { N }_{ 0 }\times { 10 }^{ -30 } }\)
\(=\frac { 93\ g\ { mol }^{ -1 }\times 2 }{ 8.55g{ cm }^{ -3 }\times 6.02\times { 10 }^{ 23 }{ mol }^{ -1 }\times { 10 }^{ -30 } }\)
\(=3.61\times { 10 }^{ 7 }=36.1\times { 10 }^{ 6 }\)
\(\therefore a={ \left( 36.1 \right) }^{ 1/3 }\times { 10 }^{ 2 }pm=3.304\times { 10 }^{ 2 }pm=330.4pm\)
\(\\ [x={ \left( 36.1 \right) }^{ 1/3 },\quad logx=\frac { 1 }{ 3 } \ log36.1=\frac { 1 }{ 3 } \times 1.5575=0.519 \ or \ x=antilog \ 0.519=3.304]\)
\(\ For \ body \ centered \ cubic,\ r=\frac { \sqrt { 3 } }{ 4 } a=0.433a\)
\( \\ =0.433\times 330.4pm=143.1 \ pm\)

\(Molarity \ (M)=\frac{\text { Mass of solute } / \text { molar mass }}{\text { Volume of solution in litres }} \\\)
M = 0.15 M = 0.15 mol L^-1 ;
Molar mass of solute = 7 x 12 + 6 x 1 x 2 x 16 = 122 g mol^-1;
Volume of solution = 250 mL = 0.25 L.
\(\left(0 \cdot 15 \mathrm{~mol} \mathrm{~L}^{-1}\right)=\frac{\text { Mass of solute }}{\left(122 \mathrm{~g} \mathrm{~mol}^{-1}\right) \times(0 \cdot 25 \mathrm{~L})}\)
Mass of solute = (0.15 mol L^-1) x (122 g mol^-1) x (0.25 L) = 4.575 g