Determine the formal charges on the highlighted atoms. if an atom is formally neutral, indicate a charge of zero. the molecule h 3 c c n o. the second c is connected to the c h 3 group with a single bond and to n with a triple bond. n is highlighted and is connected to c with a triple bond and o with a single bond. o is highlighted and is connected to n with a single bond and has 3 lone pairs of electrons.

Formal charge on n is +1

Formal charge on o is -1

Explanation:

Consider the atom n,

n is connected to o with a single bond and to c with a triple bond

In case of n, number of valence electrons are 5

Number of non-bonding electrons are 0

Number of bonding electrons are 8 (∵ n is attached to 4 bonds and each bond is equivalent to 2 electrons)

∴ Formal charge on n = 5 - 0 - (8 ÷ 2) = 5 - 4 = +1

∴ Formal charge on n is +1

Consider the atom o,

o is connected to n by a single bond

In case of o, number of valence electrons are 6

Number of non-bonding electrons are 6 (∵ it has 3 lone pair of electrons)

Number of bonding electrons are 2 (∵ it is connected to n by a single bond and that bond is equivalent to 2 electrons)

Formal charge on atom o = 6 - 6 - (2 ÷ 2) = -1

∴ Formal charge on o is -1

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the   per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:                   1 - α               α               α           = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

=  

or,

or,   = (at v = 1l)

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:               (1 - α)/0.5           α/0.5         α/0.5       = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

or,  

=  

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%