Answer:
A carbon–oxygen bond is a polar convalescent bond between carbon and oxygen. Oxygen has 6 valence electrons and prefers to either share two electrons in bonding with carbon, leaving the 4 nonbinding electrons in 2 lone pairs :O: or to share two pairs of electrons to form the carbon functional group.
Answer
is: 0.375 moles are present in 8.4 liters of nitrous oxide at stp.
V(N₂O) = 8.4 L.
V(N₂O) =
n(N₂O) · Vm.
Vm = 22,4 L/mol.<span>
n</span>(N₂O) = V(N₂O) ÷ Vm.
n(N₂O) = 8.4 L ÷ 22.4 L/mol.
n(N₂O) = 0.375 mol.<span>
Vm - molare volume on STP.</span>
The shape of XeO₂F₂ is Trigonal bi-pyramidal see-saw tetrahedron (see attached pictures)
- As you said the hybridization of Xe here is sp³d so its geometry has to be Trigonal bi-pyramidal in which F atom located on axial positions but for the final shape we exclude lone pair on Xe to give
see-saw shape (see second picture)
- Remember that we have 5 pairs (4 bond pairs + 1 lone pair) and we have to place lone pair at equatorial position.