The hydrogen atom is not actually electronegative enough to form bonds to xenon. were the xenon-hydrogen bond to exist, what wou
ld be the structure of xeh4? double-click any atom and type xe to change the label. draw the molecule by placing atoms on the grid and connecting them with bonds. show all lone pairs of electrons.
1 answer:
Answer : The structure of will be square-planar.
Explanation :
In the given molecule , 'Xe' is the central atom and 'H' is the terminal atom.
Xenon has 8 valence electrons and hydrogen has 1 valence electron. Therefore, the total number of valence electrons are 8 + 4(1) = 12 electrons.
The number of electrons used in Xe-H bonding = 8 electrons
The remaining electrons which are used as lone pair on central atom (Xe) = 12 - 8 = 4 electrons
There are 4 bonding pairs and 2 lone pairs of electrons, they will be arranged in the octahedral arrangement around the central atom with 2 lone pairs of electrons on central atom. The lone pairs are arranged linearly across the central atom. The resulting structure will be square-planar.
The structure of is shown below.
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Remark
The given thing on the right is a positron. The mass for these subatomic particles is considered to be 0. It's atomic number is 1 which means it is a blood relative of a proton.
So essentially what happens is that X is one space to the left on the periodic table. But let's solve this a little bit more formally.
Solution
y stays the same at 147. It is z that changes.
65 = z + 1 Subtract 1 from both sides.
64 = z
So the chemical with 64 as its position on the periodic table is
Gadolinium and the answer is C