One of the excited states of a c2 molecule has the electron configuration: (σ1s)2(σ1s*)2(σ2s)2(σ2s*)1(π 2p)4(σ2p)1.? what is the bond order of c2 in this excited state? what is the bond order of c2 in the ground state? how does the bond length in this excited state compare to that in the ground state?

Excited state bond order = 3

Ground state bond order = 2

Bond length excited state < bond length ground state

Explanation:

a) Bond order in excited state:

The given excited state configuration for C2 is:

(σ1s)²(σ1s*)²(σ2s)²(σ2s*)¹(π 2p)⁴(σ2p)¹

Here '*' denote antibonding orbitals

The number of electrons in bonding orbitals are 9

Number of electrons in antibonding orbitals are 3

b) Bond order in ground state

The ground state configuration for C2 is:

(σ1s)²(σ1s*)²(σ2s)²(σ2s*)²(π 2p)⁴

Here '*' denote antibonding orbitals

The number of electrons in bonding orbitals are 8

Number of electrons in antibonding orbitals are 4

c) Comparison of bond length

Bond length is inversely proportional to the bond order.

Higher the bond order lower is the bond strength and vice versa

For C2:

The ground state bond order = 2

Excited state bond order = 3

Therefore, bond length in the excited state is smaller than that in the ground state

the answer for sodium is 0.186

the answer for rubidium is 0.251

c-14 is carbon-14, or radiocarbon.   it is a radioactive isotope of carbon with an atomic nucleus containing 6 protons and 8 neutrons.