Complete the following nuclear bombardment equation by filling in the nuclear symbol for the missing species. ²³⁵₉₂U + ¹₀n → ? + ¹³⁹₅₄Xe + 3 ¹₀n

The missing species in the equation is ⁹⁴₃₈Sr

Let the missing species be ᵇₐY

We can write the equation as:

²³⁵₉₂U + ¹₀n → ᵇₐY + ¹³⁹₅₄Xe + 3¹₀n

235 + 1 = b + 139 + 3(1)

236 = b + 139 + 3

236 = b + 142

Collect like terms

236 – 142 = b

92 + 0 = a + 54 + 0

92 = a + 54

Collect like terms

92 – 54 = a

b = 94

a = 38

ᵇₐY => ⁹⁴₃₈Y => ⁹⁴₃₈Sr

The balanced equation is:

²³⁵₉₂U + ¹₀n → ⁹⁴₃₈Sr + ¹³⁹₅₄Xe + 3¹₀n

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Strontium Sr

Explanation:

By the law of mass conversion the nuclear mass remains constant in a reaction which means initial mass and atoic number will be equal to final atomic number and atomic mass respectively.

²³⁵₉₂U + ¹₀n → ? + ¹³⁹₅₄Xe + 3 ¹₀n

Here atomic mass =

235 + 1 = A + 139+3 ( A -= atomic mass of ?)

A = 236 - 142

A = 94

And, atomic number would be =

92 = B + 54

B = 92 - 54

B = 38

So, the atomic number of ? will be 38 and atomic mass will be = 94

The element with 38 atomic number is = Strontium Sr

the electronic configuration of an atom with eight electrons is 1s²2s²2p⁴. the orbital diagrams are shown in the figure.

as per the hunds rule:

1. electrons tend to avoid being in the same orbital as per as practicable.

2. electrons in different orbitals of the same energy will have their spin parallel.

this can be illustrated from the multiplicity of the orbitals of the atom. as there are four electrons in the p orbital and p orbital has 3 sub-shells. the electrons can be fed in different ways as shown in the figure but the correct orbital diagram will be as per the hunds rule.

it means there are 3 atoms of oxygen