When crystals of lead nitrate are heated strongly in a dry test tube

The correct option is D

Yellow residue is left

Explanation for the correct option

(D) Yellow residue is left

• When lead nitrate is heated vigorously in a dry test tube, lead oxide $\left(\mathrm{PbO}\right)$, nitrogen dioxide $\left({\mathrm{NO}}_2\right)$, and oxygen$\left({\mathrm{O}}_2\right)$ are produced.
• $2\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\left(\mathrm{s}\right)\stackrel{∆}{\to }2\mathrm{PbO}\left(\mathrm{s}\right)+2{\mathrm{NO}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)$
  A decomposition reaction takes place when lead nitrate crystals are heated.
• When heated, lead(II) nitrate decomposes into yellow-coloured lead (II) oxide, colourless oxygen gas, and brown-coloured nitrogen dioxide.

Hence, the correct option is (D).

Explanation for the incorrect options

(A) Crystals immediately melt.

When lead nitrate is heated lead oxide will be formed.

(B) Brown residue is left.

When lead nitrate is heated, a yellow residue is formed not a brown residue.

(C) White fumes appear in the tube.

Reddish yellow gas will be evolved on heating lead nitrate no white fumes will be formed.

Hence, the correct option is (D) Yellow residue is left