WAEC - Chemistry (2024 - No. 5)
The oxidation number of Chromium in Na\(_2\)Cr\(_2\)O\(_7\) is
+2
+6
+7
+12
Explicació
Oxidation number of Cr in Na\(_2\)Cr\(_2\)O\(_7\) = 0
2(Na) + 2Cr + 7(O) = 0
Oxidation state of Na = +1, O = - 2
So, 2(+1) + 2Cr + 7(-2) = 0
+2 + 2Cr - 14 = 0
+2 - 14 + 2Cr = 0
- 12 + 2Cr = 0
2Cr = 12
Cr = \(\frac{12}{2}\)
Cr = + 6