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in XeOF4 let x be the total number of bond pairs around Xe y be the total number of lone pairs and z be the total number of 90° angles with respect to angle OXeF then x+y+z isa)21b)25c)28d)26
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in XeOF4 let x be the total number of bond pairs around Xe y be the to...
Analysis of XeOF4 molecule:

Bond pairs around Xe (x):
In XeOF4, xenon (Xe) has 4 oxygen (O) atoms bonded to it, resulting in 4 bond pairs. Therefore, x = 4.

Lone pairs around Xe (y):
Since Xe is in group 8 (18) of the periodic table, it has 8 valence electrons. In XeOF4, each fluorine (F) atom contributes one electron to form a bond with Xe. So, the total electrons used in bonding are 4 (from O atoms) + 4 (from F atoms) = 8. This leaves Xe with 8 - 8 = 0 lone pairs. Therefore, y = 0.

90° angles with respect to angle OXeF (z):
In XeOF4, each of the 4 oxygen atoms bonded to Xe is positioned at 90° angles with respect to each other. Therefore, z = 4.

Total sum of x, y, and z:
Adding the values of x, y, and z:
x + y + z = 4 + 0 + 4 = 8
Therefore, the total sum of bond pairs, lone pairs, and 90° angles in XeOF4 is 8.

Final Answer:
The correct choice is (d) 8.
Community Answer
in XeOF4 let x be the total number of bond pairs around Xe y be the to...
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in XeOF4 let x be the total number of bond pairs around Xe y be the total number of lone pairs and z be the total number of 90° angles with respect to angle OXeF then x+y+z isa)21b)25c)28d)26
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