Class 11th

₃₇Rb = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 5s¹

Last electron enters in 5s subshell

Value of quantum numbers

n = 5, l = 0, m = 0, s =  $\pm \frac{1}{2}$

Please find the solution below:

After balancing change in oxidation state,

2Cl₂ $\stackrel{}{\to }$ 2Cl^– + 2ClO^–

Next, balance 'O' atoms,

2Cl₂ +4OH^– $\stackrel{}{\to }$  2Cl^– + 2ClO^–+ 2H₂O

Simplifying to get the simplest ratios,

Cl₂ +2OH^– $\stackrel{}{\to }$ Cl^–+ClO^–+ H₂O

x = 1, y = 2, z = 1, p = 1

Angular momentum is an axial vector along axis of rotation.

Proteins on hydrolysis gives α-amino acid because α-amino acids are building block of proteins. It is also fact that amino acids contain both –NH₂ and –COOH group.

Electronegativity generally decreases as we move down the group but Pb has higher electronegativity than Sn.

C non-metal

Si and Ge->metalloids

Sn and Pb->metals

E.N. of Sn 1.8, Pb = 1.9

(A) Fluorspar                     –    CaF₂

(B) Cryolite                 –    Na₃ [AlF₆]

(C) Bauxite                 –    Al₂O₃.2H₂O

(D) Dolomite              –    MgCO₃.CaCO₃

$p_{H_2}=P_T{\chi }_{H_2}$  (P_T = total pressure)

(  ${\chi }_{H_2}$ = mole fraction of H₂)

$p_{O_2}=P_T{\chi }_{O_2}$ ( ${\chi }_{O_2}$  = mole fraction of O₂)

$\frac{p_{H_2}}{p_{O_2}}=\frac{{\chi }_{H_2}}{{\chi }_{O_2}}=\frac{n_{H_2}}{n_{O_2}}$

$n_{H_2}=\frac{1}{2}$  mol

$n_{O_2}=\frac{1}{32}$
$\frac{p_{H_2}}{p_{O_2}}=\frac{1}{2*1}=32$

$\frac{p_{H_2}}{p_{O_2}}=\frac{32}{2}=\frac{16}{1}$