To determine the number of neutrons emitted during the nuclear fission of $${}_{92}^{235}U$$ resulting in the products $${}_{54}^{142}Xe$$ and $${}_{38}^{90}Sr$$, we need to ensure the conservation of mass number and atomic number.

The mass number (A) and atomic number (Z) have to be conserved. This means that the sum of the mass numbers and atomic numbers of the products (including any neutrons emitted) must equal those of the uranium nucleus undergoing fission.

Let's start by writing down the conservation of mass number and atomic number:

1. Conservation of Mass Number:

$${235 = 142 + 90 + n \times 1}$$

Here, $$n$$ represents the number of neutrons released. We can now calculate $$n$$:

$$ n = 235 - (142 + 90) = 235 - 232 = 3 $$

1. Conservation of Atomic Number:

$${92 = 54 + 38 + 0 \times n}$$

Neutrons do not contribute to the atomic number as they have no charge.

This calculation shows that three neutrons are needed to satisfy the conservation of mass number. The atomic number conservation also coincides, as neutrons do not alter it. Therefore, the number of neutrons emitted during this fission process is $$3$$.