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Here is my work:

Concentrated $\ce{HNO3}$ should first decompose into $\ce{NO2}$, $\ce{H2O}$ and $\ce{[O]}$. Nascent oxygen should combine with $\ce{Pb}$ to form $\ce{PbO}$. Now $\ce{PbO}$ should neutralize acid to form $\ce{Pb(NO3)2}$.

Therefore, the net products would be lead nitrate, $\ce{NO2}$ and $\ce{H2O}$.

Is this right?

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Well, you were going along the right track.... just one correction. The products will be Pb(NO₃)₂, NO₂ and H₂O. Nitric acid is a very powerful oxidizing agent. It oxidized H₂ to H₂O.

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    $\begingroup$ Oh! I forgot to add O. I have added it now. $\endgroup$ – Aumkaar Pranav Apr 11 '17 at 14:10
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Lead when reacted with nitric acid forms lead(II) nitrate, nitrogen oxide and water.

According to Wikipedia:

Lead(II) nitrate can be obtained by dissolving metallic lead in aqueous nitric acid:

$$\ce{Pb + 4 HNO3 → Pb(NO3)2 + 2 NO2 + 2 H2O}$$

Since the solvent is concentrated nitric acid (in which lead(II) nitrate has very low solubility) and the resulting solution contains nitrate ions, anhydrous crystals of lead(II) nitrate spontaneously form as a result of the common ion effect.

According to webelements:

Lead reacts slowly nitric acid, $\ce{HNO3}$. Nitrogen oxides are formed together with lead(II) nitrate, $\ce{Pb(NO3)2}$.

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The mechanics of the reaction on Pb with concentrated HNO3 likely commences as follows:

$\ce{Pb -> Pb(II) + 2 e-}$

$\ce{HNO3 = H+ + NO3-}$

$\ce{Pb(II) + 2 NO3- = Pb(NO3)2}$

$\ce{H+ + e- = .H}$

$\ce{.H + NO3- = OH- + .NO2}$ (1997 Source)

The literature also contains citations (see, for example, Page 37) of an associated mechanism involving autocatalytic $\ce{.NO2}$:

$\ce{.NO2 +e- = NO2-}$

$\ce{H+ + NO2- = HNO2}$

$\ce{HNO2 +HNO3 = H2O +2 .NO2}$

where the last reaction, I claim, can also proceeds in the reverse direction as follows:

$\ce{.NO2 + .NO2 = N2O4}$

$\ce{N2O4 + H2O = HNO2 + HNO3}$

Another 2019 paper, further cites the radical reactions:

$\ce{.H + .NO2 = .OH + .NO}$

And:

$\ce{.OH + .NO = HONO}$

One could interpret the $\ce{.NO}$ formation reaction above to imply that in dilute solutions, more of an aqueous $\ce{.NO2}$ presence and therefrom $\ce{.NO}$. Alternately, $\ce{.NO}$ could be sourced from the decomposition of any created $\ce{HNO2}$ (including from aqueous $\ce{.NO2}$) per the reaction:

$\ce{HNO2 + H+ + e− ⇌ NO + H2O}$ $\ce{ Eo = +0.98 V}$

Reference Wikipedia

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