Does lead carry the negative charge in plumbane?

Question

It says in Wikipedia that lead tetrahydride or hydrogen plumbide or plumbane ($\ce{PbH4}$) is supposed to have lead in its -4 state and the hydrogen is in its +1 state because supposedly lead has a higher electronegativity than hydrogen. Is this possible to be synthesized and actually be stable for long enough to at least view it?

Answer 1

As for experimental synthesis of plumbane, it most certainly has been done by Krivstun et al.[1], preceded by Jin and Taga[2]. The plumbane so obtained was stable enough to allow IR spectroscopy in [1] and chemical analysis in [2]. From Ref. [1]:

For the hydride generation of $\ce{PbH4}$ , we used the scheme similar to that described in [2]. Figure 1 shows a scheme of the laboratory setup for synthesis of plumbane. The basic reagents were a 5% solution of $\ce{NaBH4}$ in 0.1 g-equiv./l $\ce{Ba(OH)2}$ and a solution of lead nitrate in 0.7 M $\ce{HNO3}$ . To increase the yield of $\ce{PbH4}$ , a $\ce{K2S2O8}$ oxidizer was added to the nitrous acid solution to saturate the solution at 10°C. Lead nitrate was added to the solution directly before the performance of the reaction. The reagents were mixed in a reactor R, which represented five coils of a glass tube with inner diameter of 5 mm. The mixing zone of reagents and a spiral reactor were cooled by means of a water jacket at T = 5–10°C. The reaction products were directed to a gas–liquid separator V3 from which $\ce{PbH4}$ was carried out by the $\ce{H2}$ flow, which was produced in excess during the reaction. Then, gaseous products of the reaction were directed to a flow gas cell through a trap V4 cooled to T = –100°C. The trap served to decrease the amount of water vapors filling a measuring cell.

Special attention was paid to the uniform supply of reagents to the reaction zone and the absence of bubbles and solid particles, because in the case of deviation in the ratio of components (the excess of $\ce{NaBH4}$), reduced elementary lead appears, which causes fast decomposition of $\ce{PbH4}$. A contact of reaction products with metals and grease also results in the fast decomposition of $\ce{PbH4}$. We determined the amount of prepared $\ce{PbH4}$ by measuring the mass of Pb carried out by the $\ce{H2}$ flow from the reactor. For this purpose, we used traps V5 and V6 cooled with liquid nitrogen. The traps were evacuated after termination of the reaction and then were heated to room temperature, resulting in a complete decomposition of collected $\ce{PbH4}$. The instant of the decomposition could be observed visually by the sharp formation of an opaque black film of amorphous lead on the trap walls, which were transparent at low temperature. The amount of the lead formed was measured be the chemical method and was 3.2 mg per 1 g of $\ce{NaBH4}$ under optimum conditions of the synthesis.

References

1. Krivtsun, V. M.; Kuritsyn, Y. A.; Snegirev, E. P. (1999). "Observation of IR absorption spectra of the unstable PbH4 molecule" Opt. Spectrosc. 86 (5): 686–691.

2. Jin, K. and Taga, M. (1982). "Determination of lead by continuous-flow hydride generation and atomic absorption spectrometry". Anal. Chim. Acta 143: 229.

Answer 2

$\ce{Pb^{4+}}$ is very unstable due to the inert pair effect, wherein for p-block elements lower oxidation state becomes stabler as we move down the group. Also, the main factor is the electronegativity of lead being more than hydrogen, which you stated in your question (although they are very close). Hence the oxidation states would not be flipped.