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S
Satyajeet Singh

Beginner-Level 4

Yes, the Regional Institute of Education Common Entrance Examination (RIE CEE) 2025 for B.Ed. admissions may conduct a second and a third round of counselling depending on the availability of vacant seats after the first round. Generally, the second list or round is released a few days after the first list, followed by a waiting list round if seats remain unfilled. The official schedule and further rounds are announced separately by NCERT or the respective RIEs on their official websites.

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Manori Malhotra

Beginner-Level 5

The Union Public Service Commission will conduct the UPSC Civil Service (prelims) exam on May 24, 2026. The exam will be conducted in two sessions and the General Studies Paper 2 will be qualifying only.

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alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

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alok kumar singh

Contributor-Level 10

Kindly go through the solution 

 

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alok kumar singh

Contributor-Level 10

Due to deactivating group.

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alok kumar singh

Contributor-Level 10

Glucose and fructose are functional group isomers of each other

 

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alok kumar singh

Contributor-Level 10

Kindly go through the solution 

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alok kumar singh

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Kindly go through the solution

 

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alok kumar singh

Contributor-Level 10

[CrF6]–4

⇒ Cr+2 → 3d4

F– → WFL → No pairing so unpaired e = 4

(b)  [MnF6]–4

Mn+2 → 3d5

F– → WFL → No pairing unpaired e– = 5

(c)  [Cr (CN)6]–4 ⇒ Cr+2 ⇒ d4 CN → SFL

→ unpaired e– = 2

(d)  [Mn (CN)6]–4 ⇒ 3d5,

unpaired e– = 1

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alok kumar singh

Contributor-Level 10

[V (CO)6]

EAN = 23 – 0 + 2 * 6

= 23 + 12

= 35 ≠ 36

⇒ so it does not obey EAN rule.

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alok kumar singh

Contributor-Level 10

I2 + conc. HNO3 → HIO3 + NO2

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alok kumar singh

Contributor-Level 10

H2S2O6

→ Dithionic acid

⇒ There is no S–O–S bond in it.

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alok kumar singh

Contributor-Level 10

Acidic nature of hydrides increases down the group in p-block

so H2Te will be most acidic among given options.

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alok kumar singh

Contributor-Level 10

A jump is seen after 2nd Ip so Ve = 2

hence configuration would be ns2

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alok kumar singh

Contributor-Level 10

SO3 (s) exist as cyclic trimer ⇒ S3O9

→ sp3, so all bond π–bonds would be of pπ–dπ so pπ–pπ bond = 0

pπ–dπ bonds = 6

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alok kumar singh

Contributor-Level 10

Kindly go through the solution 

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