The problem asks us to identify a pair of molecules, where one molecule has an odd number of electrons (a free radical), and the other has an expanded octet.
Let's examine the given options:
- Option A: BeCl₂ & HNO₃ - BeCl₂ has no unpaired electrons (Be is in Group 2 and typically forms stable molecules with paired electrons). - HNO₃ is a stable molecule with no free radicals and does not have an expanded octet. Hence, this pair doesn't satisfy the given conditions.
- Option B: NO & PF₅ - NO (Nitric oxide) is a free radical with an odd number of electrons, specifically one unpaired electron. - PF₅ (Phosphorus pentafluoride) has an expanded octet. Phosphorus can accommodate more than 8 electrons because it is in Period 3 and can use d-orbitals for bonding. This option satisfies the condition of one molecule with an odd electron and the other with an expanded octet.
- Option C: BCl₃ & NO₂ - BCl₃ has no free radicals and follows the octet rule (boron typically forms 3 bonds with no lone pairs). - NO₂ is a molecule with an odd electron (it has one unpaired electron and is a free radical). However, it does not have an expanded octet. Thus, this option doesn't fit the required condition.
- Option D: SCl₂ & NH₃ - SCl₂ follows the octet rule and has no free radicals. - NH₃ (Ammonia) follows the octet rule and also has no free radicals. Hence, this option does not meet the criteria. Therefore, the correct pair is NO & PF₅, which fits the condition of one molecule having an odd electron and the other having an expanded octet.
A solid cylinder of mass 2 kg and radius 0.2 m is rotating about its own axis without friction with angular velocity 5 rad/s. A particle of mass 1 kg moving with a velocity of 5 m/s strikes the cylinder and sticks to it as shown in figure.
The angular velocity of the system after the particle sticks to it will be: