The oxidation state of P in $H_{3} P O_{4}$ is :

Question

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Answer 1

We know that oxidation state of $O = - 2, H = + 1$

Now, let's assume oxidation state of $P = x$

So, $x + 4 \times (- 2) + 3 \times (+ 1) = 0$
$\Rightarrow x - 8 + 3 = 0$
$\Rightarrow x = + 5$

Hence the oxidation state of P in $H_{3} P O_{4}$ is $+ 5$ .

Therefore option $(B)$ is correct.

Answer 2

We know that oxidation state of

O = - 2, H = + 1

Now, let's assume oxidation state of

P = x

So,

x + 4 \times (- 2) + 3 \times (+ 1) = 0

\Rightarrow x - 8 + 3 = 0

\Rightarrow x = + 5

Hence the oxidation state of P in

H_{3} P O_{4}

is

+ 5

.

Therefore option

(B)

is correct.

The oxidation state of P in $H_{3} P O_{4}$ is :

+ 3

+ 5

+ 7

None of these

We know that oxidation state of $O = - 2, H = + 1$

Now, let's assume oxidation state of $P = x$

So, $x + 4 \times (- 2) + 3 \times (+ 1) = 0$
$\Rightarrow x - 8 + 3 = 0$
$\Rightarrow x = + 5$

Hence the oxidation state of P in $H_{3} P O_{4}$ is $+ 5$ .

Therefore option $(B)$ is correct.

Oxidation state of S in $H_{2} S O_{5}$ and $H_{2} S_{2} O_{8}$ are respectively.

Statement-1 : Oxyacids of group VI A elements : $H_{2} S O 4, H_{2} S e O_{4}, H_{6} T e O_{6}$ .

Statement-2 : The lighter member where $O (2 p) \to X (3 d, 4 d) π$ bonding is significant forms strong $π - b o n d s$ to give $X O_{4}^{2 -}$ .

Tin $(S n)$ exhibits $+ 2$ and $+ 4$ oxidation states only and not $+ 3$ oxidation state.

Inert pair effect of electron pair in s-orbital is responsible.

Tin $(S n)$ exhibits $+ 2$ and $+ 4$ oxidation states only and not $+ 3$ oxidation state.

Consider the following molecules :
(I) $H_{2} O$
(II) $H_{2} S$
(III) $H_{2} S e$
(IV) $H_{2} T e$
Arrange these molecules in increasing order of bond angle.

Among the following which elements have its lower oxidation state more stable than their higher oxidation state?

Revise with Concepts

Oxoacids of Phosphorus

The oxidation state of P in H3​PO4​ is :

+ 3

+ 5

+ 7

None of these

We know that oxidation state of O=−2,H=+1 Now, let's assume oxidation state of P=x So, x+4×(−2)+3×(+1)=0⇒x−8+3=0⇒x=+5 Hence the oxidation state of P in H3​PO4​ is +5. Therefore option (B) is correct.

Oxidation state of S in H2​SO5​ and H2​S2​O8​ are respectively.

Statement-1 : Oxyacids of group VI A elements : H2​SO4,H2​SeO4​,H6​TeO6​.

Statement-2 : The lighter member where O(2p)→X(3d,4d)π bonding is significant forms strong π−bonds to give XO42−​.

Tin (Sn) exhibits +2 and +4 oxidation states only and not +3 oxidation state.

Inert pair effect of electron pair in s-orbital is responsible.

Tin (Sn) exhibits +2 and +4 oxidation states only and not +3 oxidation state.

Consider the following molecules : (I) H2​O (II) H2​S (III) H2​Se (IV) H2​Te Arrange these molecules in increasing order of bond angle.

Among the following which elements have its lower oxidation state more stable than their higher oxidation state?

Revise with Concepts

Oxoacids of Phosphorus

The oxidation state of P in $H_{3} P O_{4}$ is :

We know that oxidation state of $O = - 2, H = + 1$

Now, let's assume oxidation state of $P = x$

So, $x + 4 \times (- 2) + 3 \times (+ 1) = 0$
$\Rightarrow x - 8 + 3 = 0$
$\Rightarrow x = + 5$

Hence the oxidation state of P in $H_{3} P O_{4}$ is $+ 5$ .

Therefore option $(B)$ is correct.

Oxidation state of S in $H_{2} S O_{5}$ and $H_{2} S_{2} O_{8}$ are respectively.

Statement-1 : Oxyacids of group VI A elements : $H_{2} S O 4, H_{2} S e O_{4}, H_{6} T e O_{6}$ .

Statement-2 : The lighter member where $O (2 p) \to X (3 d, 4 d) π$ bonding is significant forms strong $π - b o n d s$ to give $X O_{4}^{2 -}$ .

Tin $(S n)$ exhibits $+ 2$ and $+ 4$ oxidation states only and not $+ 3$ oxidation state.

Tin $(S n)$ exhibits $+ 2$ and $+ 4$ oxidation states only and not $+ 3$ oxidation state.

Consider the following molecules :
(I) $H_{2} O$
(II) $H_{2} S$
(III) $H_{2} S e$
(IV) $H_{2} T e$
Arrange these molecules in increasing order of bond angle.