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Can an isotope have a charge, i.e. the number of neutrons and electrons differ in that atom?

If it does exists, is it correct to call it an ion isotope or isotope ion?

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    $\begingroup$ Yes and yes. Isotope is just an atom; it does everything that normal atom does. $\endgroup$ Jun 12 '16 at 7:08
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    $\begingroup$ In good old table salt, about 75% of the anions are $\ce{^{35}Cl^{-}}$ while 25% are $\ce{^{37}Cl^{-}}$. Nothing too special. $\endgroup$ Jun 12 '16 at 7:12
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Short answer: Yes.

Longer answer:

Isotopes are atoms of the same element that have different numbers of neutrons in their nuclei. Everything else about them is the same. The numbers of protons and electrons in a neutral atom is governed by the atomic number.

So, for example, chlorine has two naturally occurring isotopes:

$$\begin{array}{|c|c|c|c|c|c|c|}\hline \mathrm{isotope} & Z & A & p & n & e & \mathrm{abundance}\\ \hline \ce{^35_17Cl} & 17 & 35 & 17 & 18 & 17 & 75.78\%\\ \ce{^37_17Cl} & 17 & 37 & 17 & 20 & 17 & 24.22\%\\ \hline \end{array}$$ Where: $$\begin{aligned} &Z=\mathrm{atomic\ number}\\ &A = \mathrm{mass\ number}\\ &p = \mathrm{number\ of\ protons}\\ &n = \mathrm{number\ of\ neutrons}\\ &e = \mathrm{number\ of\ electrons}\\ \end{aligned}$$

Isotopes generally behave the same chemically1 since almost all chemistry is governed by the number of electrons ($Z$). Thus, anything that contains chloride ions contains a roughly 3:1 ratio of $\ce{^35Cl-}$ and $\ce{^37Cl-}$. We could call these ion isotopes or isotope ions, but in most all cases, we do not distinguish between the two in reactions or compounds involving chloride.

1 There is one caveat in the "isotopes behave the same chemically" rule of thumb that is worth mentioning. Since isotopes have different masses, they can produce kinetic isotope effects. Given the $KE=\frac{1}{2}mv^2$, two isotopes with different masses will move at different velocities and have the same kinetic energy. Heavier isotopes generally move more slowly.

This effect is most pronounced in hydrogen, where the second most common isotope $\ce{^2H}$ ($\ce{D}$) is twice as massive at the most common isotope $\ce{^1H}$ ($\ce{H}$). If two atoms have the same kinetic energy, then:

$$\begin{aligned} m_1 v_1^2 &=m_2 v_2^2\\ \dfrac{m_1}{m_2} &=\dfrac{v_2^2}{v_1^2}\\ \sqrt{\dfrac{m_1}{m_2}} &=\dfrac{v_2}{v_1} \end{aligned}$$

For the isotopes of hydrogen, the heavier isotope of hydrogen moves at only 70% of the velocity of the lighter isotope. This kinetic isotope effect can be used to probe reaction mechanisms. If the rate decreases dramatically when $\ce{H}$ is swapped for $\ce{D}$, then that hydrogen atom is involved in the rate determining step.

$$\begin{array}{|c|c|c|c|}\hline \mathrm{isotope} & A & \mathrm{rel.\ mass} & \mathrm{rel.\ velocity}\\ \hline \ce{^1H} & 1 & 1 & 1 \\ \ce{^2H} & 2 & 2 & \dfrac{1}{\sqrt{2}}\approx 0.707\\ \hline \end{array}$$

For chlorine, the outcome is a little different. Since the ratio of the masses of the two isotopes is close to 1, the ratio of the velocities is also close to 1.

$$\begin{array}{|c|c|c|c|}\hline \mathrm{isotope} & A & \mathrm{rel.\ mass} & \mathrm{rel.\ velocity}\\ \hline \ce{^35Cl} & 35 & 1 & 1 \\ \ce{^37Cl} & 37 & 1.06 & 0.97\\ \hline \end{array}$$

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  • $\begingroup$ Hmm... The origin of kinetic isotope effects lies in potential energy, or zero-point energies, not kinetic energy. The relevant quantity is not the mass $m$ of the atom in question, but the reduced mass $\mu = m_1m_2/(m_1+m_2)$ of a bond being broken. $\endgroup$
    – orthocresol
    Jan 13 at 12:31

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