Oxidation states of ${P}$ in ${H}_4 {P}_2 {O}_5, {H}_4 {P}_2 {O}_6, {H}_4 {P}_2 {O}_7$ are respectively

JEE CHEMISTRY MCQ: P-BLOCK ELEMENTS

1 Medium

1 of 1

Application +4 | -1

Oxidation states of

{P}

{H}_4 {P}_2 {O}_5, {H}_4 {P}_2 {O}_6, {H}_4 {P}_2 {O}_7

are respectively

View answer and explanation

Correct option: A

+3,+4,+5

Option D (17.325 mm Hg) is correct because it is calculated using Raoult's law for vapor pressure lowering.

Practice All Chapter MCQs →

17^{\text {th }}

{O}_2

{O}_2

\mathrm{C}{\mathrm{N}^ - }

{\mathrm{N}_2}

\mathrm{C}{\mathrm{N}^ - },\mathrm{\;}{\mathrm{N}_2}

2 {Fe}^{3+}+{SO}_2+2 {H}_2 {O} \rightarrow 2 {Fe}^{2+}+{SO}_4^{2-}+4 {H}^{+}

{H}_2 {~S}_2 {O}_8

{S}={O}

{S}-{OH}

{O}-

{O}

{NH}_3

{H}_2 {SO}_4

{H}_2 {SO}_4