Chooser the correct option : Lead oxide `PbO` can be dissoveld in
i. `HNO_3`, ii. `H_(2)SO_(4)`

Explanation

a. Opltion (ii) si correct .
Oxalic acid on heating with `H_2 SO_4` gives a mixture of two fases : `CO_2` and `CO, H_2SO_4` acts as a dehydrating agent,

When this gasesous mixture , i.e. `CO_2` and `CO`, is passed through `KOH, CO_2` gets absorbed due to the formation of `K_2 CO_3` , wherease `CO` remains as such .
`2 KOH + CO_2 rarr k_2 CO_3 +H_2 O`
`CO` combnines with `Cl_2` to form phosgene, `COCl_2` a poisonous gas . Thus , the organic acid and the two gases evlolved onm heating with conce `H_2 SO_4` are oxalic acid . `CO_2` and `CO` .
b. option (ii) is corrcet .
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c. Option (i) is correct .
Lead oxide . `PbO`, dissolves in ` HNO_3` due to formation of lead intrate `Pb (NO_3)_2` which si soluyble .
` PbO+ 2 HNO_3 rarrr Pb (NO_3)_2 +H_2 O`
However in `HC l, H_2 SO_4` and `H_2 O`, it is insolble .
`Pb + 2HCl rarr underset (white ppt.) (PbCl_2 darr) + H_2O`
`Pb + 2HCl rarr underset (white ppt.) (PbSO_4 darr) + H_2O`
`Pb + 2HCl rarr underset ("white ppt.") (Pb(OH)_2 darr) + H_2O`
d. Option (iii) is correct .
(A) the sulphide formed by passing `H_2` s gas is black , hence it cannot be `CdBr_2` as `CdS` is yellow
`Cd^(2+) + S^(2-) rarr underset ("Yellow") (Cdsdarr)`
(A) also reacts with `SnCl_2` to give whith ppt ., wghich means (A) acts as on oxdiesing agent , oxidising
`Sn^(2+) rarr Sn^(4+)` and itself it is redcued .
`HgCl_2 + H_2S rarr underset ("Black ppt") (HgS darr) + 2 H Cl`
` HgCl_2 + SnCl_2 rarr underset ("Whits ") (Hg_2Cl_2 ) +SnCl_4`
`Hg_2Cl_2 +SnCl_2 rarr underset (Grey) ( 2 Hg) + SnCl_4` .