# Question d4d86

Jun 16, 2015

THe answer is d) $\text{_15^(32)"P}$ and $\text{_16^(32)"S}$.

#### Explanation:

In order for two atoms that belong to different elements to be isobars, the must have the same mass number, $A$.

The mass number of an atom is equal to the sum of its protons and neutrons.

$A = Z + N$, where

$Z$ - the number of protons the atom has, equal to the atomic number;
$N$ - the number of neutrons it has;

For any element, the mass number is located in the upper left of the chemical symbol, while the atomic number is located in the bottom left of the symbol.

So, all you have to do to determine which of those pairs is isobaric is to take a look at the number located in the upper left of the symbol.

For $\text{_6^(12)"C}$ and $\text{_7^(14)"N}$, you have different mass numbers, $12$ and $14$, respectively, so these two atoms are not isobaric.

The same can be said for $\text{_4^9"Be}$ and $\text{_5^(11)"B}$ ($9 \cancel{\text{=}} 11$) and for $\text{_1^1"H}$ and $\text{_1^2"D}$ ($1 \cancel{\text{=}} 2$).

However, if you take a look at the atoms in the last pair, you'll notice that their mass numbers are indeed equal, which makes them isobaric.

$\text{_15^(32)"P}$ and ""_16^(32)"S"-> 32=32 => color(green)("isobaric")#