# Question 87509

Jun 1, 2017

$\sum$ of valence electrons of elements in compound

#### Explanation:

For each element, the highest principle quantum number is the valence level of the electronic structure.

=>For Main Gp Elements => Valence Electrons = Group Number

=>For Transition Metals and Inner Transition Metals => Valence level is the highest principle quantum number of the most stable electron configuration of ground state configuration.

Valence electrons in $C {H}_{4}$
V = C + 4H = 1(4) + 4(1) = 8
$C : \left[H e\right] 2 {s}^{2} 2 p 2$ => 4 valence electrons
$H : 1 {s}^{1}$ => 1 valence electron each Hydrogen => 4 valence ${e}^{-}$

Valence electrons in ${C}_{2} {O}_{4}^{2 -}$
V = 2C + 4O + 2${e}^{-}$ = 2(4) + 4(6) + 2 = 34
(Add 1 valence electron for each - charge in anions)
$C : \left[H e\right] 2 {s}^{2} 2 {p}^{2}$ => 4 valence electrons each C => 8 valence ${e}^{-}$
$O : \left[H e\right] 2 {s}^{\text{2}} 2 {p}^{4}$ => 6 valence electrons each O => 24 valence ${e}^{-}$
$+ 2 {e}^{_}$ => excess of 2 ${e}^{-}$accounts for the -2 oxidation state

Valence electrons in $N {H}_{4}^{+}$
V = 1N + 4H - 1${e}^{-}$ 1(5) + 4(1) - 1 = 8
(Subtract 1 valence electron for each + charge in cations)
$N : \left[H e\right] 2 {s}^{\text{2}} 2 {p}^{3}$ => 5 valence electrons
$H : 1 {s}^{1}$ => 1 valence electron each Hydrogen => 4 valence ${e}^{-}$
$- 1 {e}^{-}$ => deficiency of 1 ${e}^{-}$ accounts for +1 oxidation state

Valence electrons in ${S}_{2} {O}_{3}^{2 -}$
V = 2S + 3O + 2${e}^{-}$ = 2(6) + 3(6) + 2 = 32
$S : \left[N e\right] 3 {s}^{\text{2}} 3 {p}^{4}$ => 6 valence electrons each S
=> 2(6) = 12 valence electrons
$O : \left[H e\right] 2 {s}^{\text{2}} 2 {p}^{4}$ => 6 valence electrons each Oxy
=> 3(6) = 18 valence electrons
$+ 2 {e}^{_}$ => excess of 2 ${e}^{-}$accounts for the -2 oxidation state

Valence electrons in $A g {\left(N {H}_{3}\right)}_{2}^{+}$
V = Ag + 2N + 6H - 1${e}^{-}$= 1 + 2(5) + 6(1) - 1 = 16
$A g : \left[K r\right] 4 {d}^{10} 5 {s}^{1}$ => 1 valence electron (1 ${e}^{-}$ in 5s orbital after stabilizing 4d ."orbital").
$N : \left[H e\right] 2 {s}^{\text{2}} 2 {p}^{3}$ => 5 valence electrons each Nitrogen => 2(5) = 10 valence electrons
$H : 1 {s}^{1}$ => 1 valence electron each Hydrogen => 6 valence ${e}^{-}$
$- 1 {e}^{-}$ => deficiency of 1 ${e}^{-}$ accounts for +1 oxidation state