Which of the following represents Plumbous ion?

P b^{+ 2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

P b^{+ 4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

P b^{+ 3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

P b^{+ 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $P b^{+ 2}$
Lead looses 2 electrons when combines with other element.
So Plumbous ion is written as; $P b^{2 +}$ .

Suggest Corrections

Similar questions

P b^{+ 4}

[1] P b^{2 +} > P b^{4}, S n^{2 +} > S n^{4 +}

[2] P b^{2 +} < P b^{4}, S n^{2 +} < S n^{4 +}

[3] P b^{2 +} > P b^{4}, S n^{2 +} < S n^{4 +}

[4] P b^{2 +} < P b^{4}, S n^{2 +} > S n^{4 +}

[5] S n^{2 +} < P b^{2 +}, S n^{4 +} > P b^{4 +}

[6] S n^{2 +} < P b^{2 +}, S n^{4 +} < P b^{4 +}

lead reacts with oxygen. which ion should we take for lead? pb+1 or pb+2 or pb+4....

Identify the monoatomic and polyatomic ions from the following equation-
$P b (N O_{3})_{2}$ + $K_{2} S O_{4}$ $\to$ $2 K N O_{3}$ + $P b S O_{4}$

P b^{4 +} + S n^{2 +} \to S n^{4 +} + P b^{2 +}

S n^{4 +} + P b^{2 +} \to P b^{4 +} + S n^{2 +}

E_{P b^{4 +} / P b^{2 +}}^{o} = 1.69 V, E_{S n^{4 +} / S n^{2 +}}^{o} = 0.15 V

Join BYJU'S Learning Program