Final answer:
The ions that match the nd8 electron configuration requirement are Pt2+ and Ni2+. Pt2+ has an electron configuration of [Xe]4f14 5d8, and Ni2+ would have [Ar]3d8. The other ions listed do not have an nd8 configuration after becoming ions.
Step-by-step explanation:
We are tasked with finding ions that have an electron configuration ending in nd8. The suffix 'nd8' indicates that there are eight electrons in the 'd' subshell of the 'n'th energy level. To have an nd8 electron configuration, the ion must be a d-block element (transition metal) that has lost a sufficient number of electrons to leave it with only eight electrons in its outermost d subshell.
Let's consider each option given:
- Pt2+: Platinum originally has an electron configuration of [Xe]4f14 5d9 6s1. When it loses two electrons to form the Pt2+ ion, it achieves the electron configuration [Xe]4f14 5d8, which fits the nd8 criteria.
- Ni2+: Nickel has an electron configuration of [Ar]3d8 4s2. Losing two electrons to form Ni2+ would result in a configuration of [Ar]3d8, which is also nd8.
The other options do not match the criterion for nd8 configuration after ionization:
- P3+ would have an electron configuration of [Ne], not nd8.
- Cr6+ would have a configuration of [Ar], again not nd8.
- Ag3+ would vastly exceed the common oxidation states of silver and is unlikely.
- Gd3+ would have a configuration of [Xe]4f7, not nd8.