64. Diberikan data beberapa unsur transisi periode-4 UNSUR & BILANGAN OKSIDASI mathrm(Q) & +3 mathrm(R) & +1,+2,+3 mathrm(S) & +2,+3,+4,+5 mathrm(T) & +3,+4,+5,+6,+7 Spesi yang mungkin adalah (A) mathrm(QCl)_(2) (B) mathrm(RO)_(3)^2+ mathrm(X) (C) mathrm(SH)_(2)^3 (D) mathrm(T)(mathrm(NO)_(3))_(2) (E) mathrm(TBr)_(5)( )^2-

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64. Diberikan data beberapa unsur transisi periode-4 UNSUR & BILANGAN OKSIDASI mathrm(Q) & +3 mathrm(R) & +1,+2,+3 mathrm(S) & +2,+3,+4,+5 mathrm(T) & +3,+4,+5,+6,+7 Spesi yang mungkin adalah (A) mathrm(QCl)_(2) (B) mathrm(RO)_(3)^2+ mathrm(X) (C) mathrm(SH)_(2)^3 (D) mathrm(T)(mathrm(NO)_(3))_(2) (E) mathrm(TBr)_(5)( )^2-

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Answer 1

(E) $TBr_{5}^{2-}$

Answer 2

Step 1:

Identify the possible oxidation states for each element

From the table, we know the oxidation states of each element:

- Element Q: +3
- Element R: +1, +2, +3
- Element S: +2, +3, +4, +5
- Element T: +3, +4, +5, +6, +7

Step 2:

Analyze the given species and their feasibility based on oxidation states

We need to determine if the given species can exist based on the oxidation states of the elements involved.

Option (B) $RO_{3}^{2-}$:

- For $RO_{3}^{2-}$ to be stable, R must have an oxidation state that balances the charge of the ion.
- Oxygen typically has an oxidation state of -2. Therefore, in $RO_{3}^{2-}$, the total oxidation state contributed by oxygen is $3 \times (-2) = -6$.
- To balance this, R must have an oxidation state of +4 ($+4 - 6 = -2$).
- However, R only has oxidation states of +1, +2, and +3. Thus, $RO_{3}^{2-}$ is not feasible.

Option (C) $SH_{2}$:

- For $SH_{2}$, S must have an oxidation state that balances the hydrogen atoms.
- Hydrogen typically has an oxidation state of +1. Therefore, in $SH_{2}$, the total oxidation state contributed by hydrogen is $2 \times (+1) = +2$.
- To balance this, S must have an oxidation state of -2.
- However, S only has oxidation states of +2, +3, +4, and +5. Thus, $SH_{2}$ is not feasible.

Option (D) $T(NO_{3})_{2}$:

- For $T(NO_{3})_{2}$, T must have an oxidation state that balances the nitrate ions.
- Nitrate ($NO_{3}^{-}$) has a charge of -1. Therefore, two nitrate ions contribute a total charge of $2 \times (-1) = -2$.
- To balance this, T must have an oxidation state of +2.
- Since T has oxidation states of +3, +4, +5, +6, and +7, it cannot have an oxidation state of +2. Thus, $T(NO_{3})_{2}$ is not feasible.

Option (E) $TBr_{5}^{2-}$:

- For $TBr_{5}^{2-}$, T must have an oxidation state that balances the bromine atoms and the overall charge.
- Bromine typically has an oxidation state of -1. Therefore, in $TBr_{5}^{2-}$, the total oxidation state contributed by bromine is $5 \times (-1) = -5$.
- To balance this, T must have an oxidation state of +3 ($+3 - 5 = -2$).
- Since T has an oxidation state of +3, $TBr_{5}^{2-}$ is feasible.

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