What is the IUPAC name of compound NaBH4?a)Sodium boronhydrideb)Sodium tetrahydridoboron (III)c)Sodium tetrahydridoborate (III)d)Sodium tetrahydridoborate (I)Correct answer is option 'C'. Can you explain this answer?

Question

Answer

Here boron is in +3 O.S. 4 H- atoms which are named as hydriod are in coordination sphere while Sodium is in ionization sphere. Hence the name Sodium Tetrahydridoborate(III)

Answer

It is anionic complex so the sequence is SODIUM TETRAHYDROIn general anionic ligand name ends in -ide ,-ite or -ate so BORATEThe O.S of Boron in given compound is 1+x+(-1×4)=0》x=3. So( III)

Answer

The IUPAC name of the compound NaBH4 is Sodium tetrahydridoborate (III).

Explanation:
- Sodium: The first part of the name indicates the cation present in the compound, which is sodium (Na+).
- Tetrahydridoborate: The second part of the name indicates the anion present in the compound, which is tetrahydridoborate (BH4-).

Now, let's break down the name further:

1. Sodium (Na+):
- Sodium is a metal that forms a +1 cation, represented as Na+.
- It is named as sodium according to the IUPAC convention.

2. Tetrahydridoborate (BH4-):
- Tetrahydridoborate is an anion composed of boron (B) and hydrogen (H) atoms.
- The prefix 'tetra-' indicates that there are four hydrogen atoms bonded to the boron atom.
- The suffix '-hydrido' indicates the presence of hydrogen atoms.
- The suffix '-borate' indicates that the compound contains boron.
- The overall charge of the tetrahydridoborate anion is -1, represented as BH4-.

3. Roman Numeral (III):
- The Roman numeral in parentheses indicates the oxidation state of the boron atom in the compound.
- In NaBH4, the boron atom has an oxidation state of -3, which is represented by the Roman numeral III.

Therefore, the correct IUPAC name for NaBH4 is Sodium tetrahydridoborate (III).

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