A molecule has sp3d hybridization with 3 lone pairs. ... The electron pair geometry of this molecule is

The shape of the  hybridized molecule with three lone pairs has been linear.

The VSEPR theory has been applied for the determination of the shape of the molecule with respect to the hybridization of the molecule.

According to the VSEPR theory, the molecule with hybridization and zero lone pairs has been consisted of the trigonal bipyramidal structure.

The lone pairs in the molecule result in the repulsion that distorts the shape of the molecule from trigonal bipyramidal. The resulted structure has a linear shape.

The shape of the  hybridized molecule with three lone pairs has been linear.

For more information about the hybridization, refer to the link:

link

Linear

Explanation:

Usually, when a molecule has a hybridization of sp3d, it's geometry is trigonal bipyramidal.

However, the presence of lone pairs distorts the shape away from the expected trigonal bipyramidal geometry.

The three lone pairs present in the molecule leads to a linear shape due to repulsion between electron pairs.

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

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