Draw structure `H_(2)PO_(3)^(-)`

Draw structure CsH_(2)PO_(2)

Draw structure H_(2)P_(2)O_(5)^(2-)

Lewis -structure PO_(3)^(-3)

For H_(3)PO_(4) , H_(3)PO_(4) rarr H_(2)PO_(4)^(-)+H^(+)(K_(1)), H_(2)PO_(4)^- rarr HPO_(4)^(-)+H^(+) (K_2), HPO_(4)^(2-) rarr PO_(4)^(3-) + H^(+) (K_(3)) then

Three reactions involving H_(2)PO_(4)^(-) are given below I. H_(3)PO_(4)+H_(2)OrarrH_(3)O^(+)+H_(2)PO_(4)^(-) II. H_(2)PO_(4)^(-)+H_(2)OrarrHPO_(4)^(2-)+H_(3)O^(+) III. H_(2)PO_(4)^(-)+OH^(-)rarrH_(3)PO_(4)+O^(2-) In which of the above does H_(2)PO_(4)^(-) act as an acid?

Classify the following into acid base and amphiprotic species on the basis of protonic concept : H_(2)PO_(2)^(-) (ii) H_(2)PO_(3)^(-),(III) H_(2)PO_(4)^(-),(iv) HPO_(3)^(2-),(v)HPO_(4)^(2-)(vi)NH_(4)^(+) (vii)CH_(3)COOH_(2)^(+)

In the following unbalanced redox reaction, Cu_(3)P+Cr_(2)O_(7)^(2-) rarr Cu^(2+)+H_(3)PO_(4)+Cr^(3+) Equivalent weight of H_(3)PO_(4) is

H_(3)PO_(4)+H_(2) Leftrightarrow H_(3)O^(+)+H_(2)PO_(4)^(-),pK_(1)=2.15 H_3PO_(4)^(-)+H_(2)O Leftrightarrow H_(3)O^(+)+HPO_(4)^(2-), pK_(2)=7.20 Hence pH of 0.01 M NaH_(2)PO_(4) is

Account for the following : (i) Sulphur in vapour from exhibits paramagnetic behavious. (ii) SnCl_(4) is more covalent the SnCl_(2) . (iii) H_(3)PO_(2) is a stronger reducing agent then H_(3)PO_(3) .

H_(3)PO_(4)hArr H^(o+) +H_(2)PO_(4)^(Theta), K_(a_(1)) : H_(2)PO_(4)^(Theta) hArr H^(o+) +HPO_(4)^(2-),' K_(a_(2)) : HPO_(4)^(2-) hArr H^(o+) + PO_(4)^(3-), K_(a_(3)): Mark out the incorrect statements: