assign oxidation number to the underlined elements nah2po4

Find the oxidation number of P in N a H 2 P O 4 .

Let x be the oxidation number of p in n a h 2 p o 4 . the oxidation numbers of na, h and o are + 1 , + 1 and − 2 respectively. + 1 + 2 ( + 1 ) + x + 4 ( − 2 ) = 0 3 + x + − 8 = 0 x + − 5 = 0 x = + 5 the oxidation number of p in n a h 2 p o 4 is + 5 ..

Assign oxidation numbers to the underlined elements in each of the following species:

(a) NaH 2 P O 4 (b) NaH S O 4 (c) H 4 P 2 O 7 (d) K 2 Mn O 4

(e) Ca O 2 (f) Na B H 4 (g) H 2 S 2 O 7 (h) KAl( S O 4 ) 2 .12 H 2 O

Assign oxidation number to the underlined element in N a H 2 P – – O 4 .

Oxidation number of P in M g 2 P 2 O 7 is?

Find the oxidation number of P in P 2 O 4 − 7 .

Assign oxidation numbers to the underlined elements in each of the following species: (a) NaH 2 P O 4 (b) NaH S O 4 (c) H 4 P 2 O 7 (d) K 2 Mn O 4 (e) Ca O 2 (f) Na B H 4 (g) H 2 S 2 O 7 (h) KAl( S O 4 ) 2 .12 H 2 O

(a) let the oxidation number of p be x . we know that, oxidation number of na = +1 oxidation number of h = +1 oxidation number of o = –2 then, we have hence, the oxidation number of p is +5. (b) then, we have hence, the oxidation number of s is + 6. (c) then, we have hence, the oxidation number of p is + 5. (d) then, we have hence, the oxidation number of mn is + 6. (e) then, we have hence, the oxidation number of o is – 1. (f) then, we have hence, the oxidation number of b is + 3. (g) then, we have hence, the oxidation number of s is + 6. (h) then, we have or, we can ignore the water molecule as it is a neutral molecule. then, the sum of the oxidation numbers of all atoms of the water molecule may be taken as zero. therefore, after ignoring the water molecule, we have hence, the oxidation number of s is + 6..

Assign oxidation numbers to the underlined elements in each of the following species: N a H 2 P O 4

N a H S – – O 4

H 4 P – – 2 O 7

K 2 M n – –– – O 4

C a O – – 2

N a B – – H 4

H 2 S – – 2 O 7

K A l S – – O 4 ) 2 .12 H 2 O

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Find the oxidation number of P in $Na{H_2}P{O_4}$.

Repeaters Course for NEET 2022 - 23

Calculate the oxidation number of the underlined atom. NaH2PO4 - Chemistry

Calculate the oxidation number of the underlined atom.

NaH 2 P O 4

Solution Show Solution

Oxidation number of Na = +1

Oxidation number of H = +1

Oxidation number of O = –2

NaH 2 PO 4 is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms = 0

∴ (Oxidation number of Na) + 2 × (Oxidation number of H) + (Oxidation number of P) + 4 × (Oxidation number of O) = 0

∴ (+1) + 2 × (+1) + (Oxidation number of P) + 4 × (–2) = 0

∴ (Oxidation number of P) + 3 – 8 = 0

∴ Oxidation number of P in NaH 2 PO 4 = +5

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Let expect oxidation number of P is x.

We realize that,

\[\begin{array}{*{35}{l}}

Oxidation\text{ }number\text{ }of\text{ }Na\text{ }=\text{ }+1  \\

Oxidation\text{ }number\text{ }of\text{ }H\text{ }=\text{ }+1  \\

Oxidation\text{ }number\text{ }of\text{ }O\text{ }=\text{ }-\text{ }2  \\

\end{array}\]

Then, at that point, we have

1\left( +1 \right)\text{ }+\text{ }2\left( +1 \right)\text{ }+\text{ }1\text{ }\left( x \right)\text{ }+\text{ }4\left( -\text{ }2 \right)\text{ }=\text{ }0  \\

=\text{ }1\text{ }+\text{ }2\text{ }+\text{ }x\text{ }-\text{ }8\text{ }=\text{ }0  \\

=\text{ }x\text{ }-\text{ }5\text{ }=\text{ }0  \\

=\text{ }x\text{ }=\text{ }+\text{ }5  \\

Subsequently the oxidation number of P is +5

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Assign oxidation number to the underlined element in each of the following species N a H 2 P O 4 ( b ) N a H S O 4 ( c ) H 4 P 2 O 7 ( d ) K 2 M n O 4 ( e ) C O 2 ( f ) N a B H 4 ( g ) H 2 S 2 O 7 ( h ) K A I ( S O 4 ) 2 12 H 2 O

Recommended questions, (a) let the oxidation number of p be x writing the oxidation number of each atom above its symbol we sum of oxidation numbr of various atoms in n a h 2 p o 4 = 1 ( + 1 ) + 2 ( + 1 ) + 1 ( x ) + 4 ( − 2 ) = x − 5 but the sum of oxdation number of various atoms in n a h 2 p o 4 (neutral) is zero thus the oxidation nuber of p in n a h 2 p o 4 = 5 (b) + 1 n a x h x s − 2 o 4 ∴ + 1 ( + 1 ) + x + 4 ( + 1 ) − 2 = 0 or x = 6 thus the oxidation number of s in n a h s o 4 = + 5 (c ) + 1 h 4 + 1 p 2 x s − 2 o 4 ∴ 4 ( + 1 ) + 2 ( x ) + 7 ( − 2 ) = 0 or x = + 5 (d) k 2 m n o − 2 4 ∴ 2 ( + 1 ) + 1 ( x ) + 7 ( − 2 ) = 0 or x = + 6 thus the oxidation numbr of mn in k 2 m n i o 4 = + 7 (e ) let hte oxidation number of o be x since ca is an alkaline earth metal therefore its oxidation number is +2 thus c a o 2 ∴ + 2 + 2 ( x ) = 0 or x = − 1 (f) in n a b h 4 h is present as hydride ion therefore its oxidation number ois -1 thus n a b h 4 ∴ 2 ( + 1 ) + x + ( − 1 ) = 0 or x = + 3 thus the oxidation number of b in n a b h 4 = + 3 (g) h 2 s 2 o 2 − 7 ∴ 2 ( + 1 ) + 2 ( x ) + 7 ( − 2 ) = 0 or x 6 thus the oxidation number of s in h 2 s 2 o 7 = + 6 (h) k a i ( s o 4 ) 12 ( h 2 o or + 1 + 3 + 2 x + 8 ( − 2 ) + 12 ( 2 × 1 − 2 ) or x = + 6 alternatively since h 2 o is a neutral molecule therefore sum of oxidation number of s ∴ + 1 + 3 + 2 + x − 16 = 0 or x = + 6 thus the oxidation number of s in kai ( s o 4 ) 12 h 2 o = + 6, similar questions.

Assign oxidation number to the underlined elements in each of the following species. a) NaH_(2)underline(P)O_(4)" " b) NaHunderline(S)O_(4)" " c) H_(4)underline(P_(2))O_(7) d) K_(2)underline(Mn)O_(4)" "e) Caunderline(O_(2))" " f) Naundeline(B)H_(4) g) H_(2)underline(S_(2))O_(7)" " h) KAlunderline(SO_(4))_(2).12H_(2)O

Assign oxidation number to the underlined elements in each of the following species. a) N a H 2 P – – O 4 b ) N a H S – – O 4 c ) H 4 P 2 – – – O 7 d) K 2 M n – –– – O 4 e ) C a O 2 – – – f ) N a u n ∂ ∈ e ( B ) H 4 g) H 2 S 2 – – – O 7 h ) K A l S O 4 – ––– – 2 .12 H 2 O

Balance the following equations : (i) KMnO_(4) + H_(2)SO_(4) + H_(2)O_(2) to K_(2)SO_(4) + MnSO_(4) + H_(2)O + O_(2) (ii) KMnO_(4) + KCl + H_(2)SO_(4) to MnSO_(4) + K_(2)SO_(4) + H_(2)O + Cl_(2) (iii) MnO_(2) + H_(2)O_(2) to MnO_(4)^(-) + MnO_(4)^(-) + H_(2)O (Basic medium)

Assign oxidation number to the underline elements in each of the following species: H_(4)underline (P)_(2)O_(7)

Assign oxidation number to the underline elements in each of the following species: KAIunderline((S)O_(4))(2).12H_(2)O

Assign oxidation number to the underlined elements in each of the following species: a. NaH_(2)PO_(4) b. NaHul(S)O_(4) c. H_(4)ul(P_(2))O_(7) d. K_(2)ul(Mn)O_(4) e. ul(Ca)O_(2) f. Naul(B)H_(4) g. H_(2)ul(S_(2))O_(7) h. KAl(ul(S)O_(4))_(2).12H_(2)O

Assign oxidation number to the underlined elements in each of the following species : (a) NaH_(2)underlinePO_(4) (b) NaHunderlineSO_(4) ( c) H_(4)underlineP_(2)O_(7) (d) K_(2)underline(Mn)O_(4) (e) CaunderlineO_(2) (f) NaunderlineBH_(4) (g) H_(2)underlineS_(2)O_(7) (h) KAl(underlineSO_(4))_(2)*12H_(2)O

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