Respuesta :
The oxidation state of Na, B, and H in NaBH₄ are +1, +3, and -1 respectively.
The oxidation state of Al, B, and H in Al(BH₄)₃ are +3, +3, and -1 respectively
The oxidation state of Li, Al, and H in LiAlH₄ are +1, +3, and -1 respectively
In metal hydrides, hydrogen has an oxidation state of -1.
a) NaBH₄
Let x be the boron's oxidation state.
Oxidation state of H = -1
For four hydrogen, oxidation state = -4
Oxidation state of Na = +1
+1 + x + (-4) = 0
x - 3 = 0
x = 3
Hence, the Oxidation state of B is +3
2) Al(BH₄)₃
Let the oxidation state of Boron is b
The oxidation state of Al = +3
Oxidation state of 12 H = -12
+3 + 3b + (-12) = 0
3b - 9 = 0
3b = 9
b = +3
Hence, the oxidation state of B is +3
3) LiAlH4
Let's take the oxidation state of Al = a
Oxidation state of Li = +1
Oxidation state of 4 H = -4
+1 + a +(-4) = 0
a =+3
Hence, the oxidation state of Al is +3
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