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Chapter 3: Problem 104

How many electrons are there in \(\mathrm{Zr}^{4+}\) ? What \(2+\) cation has the same number of electrons as \(\mathrm{Zr}^{4+} ?\)

Short Answer

Expert verified
There are 36 electrons in the \(\mathrm{Zr}^{4+}\) ion, and the 2+ cation with the same number of electrons is \(\mathrm{Sr}^{2+}\).

Step by step solution

01

Find the atomic number of Zirconium (Zr)

We can either look up the atomic number of Zirconium in the periodic table or remember that it is 40. So, the atomic number of Zirconium (Zr) is 40.
02

Determine the number of electrons in \(\mathrm{Zr}^{4+}\)

Since Zirconium loses 4 electrons to form the \(\mathrm{Zr}^{4+}\) cation, we can find the number of remaining electrons by subtracting 4 from 40: Number of electrons in \(\mathrm{Zr}^{4+}\) = atomic number of Zr - 4 = 40 - 4 = 36 So, there are 36 electrons in the \(\mathrm{Zr}^{4+}\) ion.
03

Find the 2+ cation with the same number of electrons

Now we have to find the element which forms a 2+ cation with the same number of electrons as in \(\mathrm{Zr}^{4+}\). We know that the 2+ cation loses 2 electrons, so we need to find an element whose atomic number minus 2 equals the number of electrons in \(\mathrm{Zr}^{4+}\), which is 36. Let's use 'X' as the atomic number of the unknown element: X - 2 = 36 X = 36 + 2 = 38 The element with atomic number 38 is Strontium (Sr). Therefore, the 2+ cation with the same number of electrons as \(\mathrm{Zr}^{4+}\) is \(\mathrm{Sr}^{2+}\).

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Most popular questions from this chapter

Two different compounds, both consisting of sodium (Na) and oxygen (O), were analyzed. The data are given below: $$ \begin{array}{cccc} & \begin{array}{l} \text { Mass of } \\ \text { sample } \\ \text { Compound } \end{array} & \begin{array}{l} \text { Mass of } \\ \text { analyzed } \end{array} & \text { O present } & \begin{array}{l} \text { Mass of } \\ \text { Na present } \end{array} \\ \hline \mathrm{A} & 19.50 \mathrm{~g} & 8.00 \mathrm{~g} & ? \\ \mathrm{~B} & 61.98 \mathrm{~g} & 16.00 \mathrm{~g} & ? \\ \hline \end{array} $$ (a) Fill in the last column of the table. (b) Calculate the \(\% \mathrm{Na}\) and \(\% \mathrm{O}\) for both compounds.

What is the identity of element \(X\) if its \(2+\) cation contains ten electrons?

Fill in the following table: $$ \begin{array}{lcccc} & { }_{8}^{15} \mathrm{O}^{+} & ? & ? & ? \\ \hline \text { Mass number } & ? & 27 & ? & 58 \\ \text { Atomic number } & ? & ? & 15 & ? \\ \text { Number of protons } & ? & 13 & ? & ? \\ \text { Number of neutrons } & ? & ? & 16 & 30 \\ \text { Number of electrons } & ? & ? & ? & 27 \\ \text { Charge on ion } & ? & +3 & -3 & +1 \\ \hline \end{array} $$

How does a group differ from a period in the periodic table?

Use the elements lithium, potassium, and sodium to explain what is meant by chemical periodicity.
See all solutions

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