Predict the chemical formulas of the compounds formed by the following pairs of ions: (a) \(\mathrm{Cr}^{3+}\) and \(\mathrm{CN}^{-}\), (b) \(\mathrm{Mn}^{2+}\) and \(\mathrm{ClO}_{4}^{-}\), (c) \(\mathrm{Na}^{+}\) and $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\(, (d) \)\mathrm{Cd}^{2+}\( and \)\mathrm{CO}_{3}^{2-}$, (e) \(\mathrm{Ti}^{4+}\) and \(\mathrm{O}^{2-}\).

Short Answer

Expert verified
The chemical formulas of the compounds formed by the given pairs of ions are: (a) \(\mathrm{Cr(CN)}_{3}\) (b) \(\mathrm{Mn(ClO}_{4})_{2}\) (c) \(\mathrm{Na}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\) (d) \(\mathrm{CdCO}_{3}\) (e) \(\mathrm{TiO}_{2}\)

Step by step solution

01

(a) \(\mathrm{Cr}^{3+}\) and \(\mathrm{CN}^{-}\)

To form a compound with a neutral charge between \(\mathrm{Cr}^{3+}\) and \(\mathrm{CN}^{-}\), we need to find the lowest common multiple (LCM) of the charges 3 and 1. The LCM is 3. This means that we need three \(\mathrm{CN}^{-}\) ions to balance one \(\mathrm{Cr}^{3+}\) ion. The chemical formula for the compound is \(\mathrm{Cr(CN)}_{3}\).
02

(b) \(\mathrm{Mn}^{2+}\) and \(\mathrm{ClO}_{4}^{-}\)

In the case of \(\mathrm{Mn}^{2+}\) and \(\mathrm{ClO}_{4}^{-}\), the LCM of 2 and 1 is 2. This means that we need two \(\mathrm{ClO}_{4}^{-}\) ions to balance one \(\mathrm{Mn}^{2+}\) ion. The chemical formula for the compound is \(\mathrm{Mn(ClO}_{4})_{2}\).
03

(c) \(\mathrm{Na}^{+}\) and \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}\)

For the ions \(\mathrm{Na}^{+}\) and \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}\), the LCM of 1 and 2 is 2. This means that we need two \(\mathrm{Na}^{+}\) ions to balance one \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}\) ion. The chemical formula for the compound is \(\mathrm{Na}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\).
04

(d) \(\mathrm{Cd}^{2+}\) and \(\mathrm{CO}_{3}^{2-}\)

In the case of \(\mathrm{Cd}^{2+}\) and \(\mathrm{CO}_{3}^{2-}\), the charges of the ions are equal and opposite, which means that we need one of each ion to balance each other out. The chemical formula for the compound is \(\mathrm{CdCO}_{3}\).
05

(e) \(\mathrm{Ti}^{4+}\) and \(\mathrm{O}^{2-}\)

Finally, for the ions \(\mathrm{Ti}^{4+}\) and \(\mathrm{O}^{2-}\), the LCM of 4 and 2 is 4. This means that we need two \(\mathrm{O}^{2-}\) ions to balance one \(\mathrm{Ti}^{4+}\) ion. The chemical formula for the compound is \(\mathrm{TiO}_{2}\).

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