Grade eleventh chemistry
▎Question 1

What does VSEPR stand for?
• A) Valence Shell Electron Pair Repulsion
• B) Valence Shell Electron Probability Repulsion
• C) Valence Shell Energy Pair Repulsion
• D) Valence Shell Electron Pair Reaction

Answer

---

▎Question 2

According to VSEPR theory, the shape of a molecule is primarily determined by:
• A) The number of protons in the nucleus
• B) The arrangement of electrons around the central atom
• C) The type of chemical bonds present
• D) The molecular weight of the compound

Answer:
---

▎Question 3

What is the molecular geometry of a molecule with the formula  AB₂  where  A  is the central atom and there are no lone pairs?
• A) Linear
• B) Bent
• C) Tetrahedral
• D) Trigonal planar

Answer:
---

▎Question 4

Which of the following molecular geometries corresponds to a molecule with the formula  AB₃E ?
• A) Tetrahedral
• B) Trigonal bipyramidal
• C) Trigonal planar
• D) Seesaw

Answe

---

▎Question 5

In a molecule represented as  AB₂E₂ , what is the arrangement of electron pairs around the central atom?
• A) Linear
• B) Tetrahedral
• C) Trigonal planar
• D) Octahedral

Answer

---

▎Question 6

For a molecule with the formula  AB₄ , what is the expected molecular shape?
• A) Square planar
• B) Tetrahedral
• C) Octahedral
• D) Linear

Answer:

---

▎Question 7

Which of the following statements is true regarding molecules with one or more lone pairs on the central atom?
• A) They always have symmetrical shapes.
• B) They can lead to distorted geometries.
• C) They cannot participate in bonding.
• D) They have no effect on molecular shape.

Answer:
---

▎Question 8

What is the molecular geometry of  AB₃E₃ ?
• A) T-shaped
• B) Octahedral
• C) Square planar
• D) Trigonal bipyramidal

Answer:

▎Question 9

When applying VSEPR theory, which of the following factors should be considered?
• A) Electronegativity differences between atoms
• B) Number of lone pairs on the central atom
• C) Molecular weight of the compound
• D) Bonding energy of the atoms involved

Answer:
---

▎Question 10

In which molecular geometry do all bond angles measure approximately 109.5 degrees?
• A) Tetrahedral
• B) Trigonal planar
• C) Linear
• D) Octahedral

Answer:

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An Ethiopian Developer Who Once Failed National Exams Built a Platform to Help Others Pass

Since exam reforms in 2022, only about 4.6% of Ethiopian students have passed national entrance exams. One ed-tech startup says the problem isn’t just content, it’s structure.
January 16, 2026
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Daniel Metaferiya

Addis Ababa, Ethiopia

Entering college has become an increasingly fraught milestone for Ethiopian students in recent years. Since reforms to the national examination system in 2022, pass rates have plunged: more than 3.1 million students have sat for the exams over the past four years, and only about 4.6 percent have passed. What was once a difficult rite of passage has become, for many, a near wall.

Facing those odds at the end of high school, students are trying a range of strategies to clear the hurdle. One education-technology startup, founded by a young software engineer who failed the exam on his first attempt, is betting that better structure, not just more study materials, can make the difference.

The platform, TikuretEntrance, was founded by Dagim Wallelgne and is designed to support students from the moment they enter high school, not just in the frantic months before the test. It offers curriculum-aligned lessons, past national exam questions, mock tests, and performance-tracking tools tailored to Ethiopia’s education system. 

“Not all students fail because they lack materials,” Dagim told Shega. “Many struggle because they lack focus, structure, and regular guidance.”

TikuretEntrance is available in English, Amharic, Afan Oromo, and Tigrinya, widening access for students across regions and language backgrounds. It covers core subjects, including mathematics, natural sciences, English, and civics, combining digital notes with timed practice exams designed to mirror real testing conditions. After each assessment, students receive automated feedback and detailed explanations, helping them identify weak areas and adjust their study plans.

“I failed the exam first, not because of academic weakness, but because I lacked structure,” Dagim recalled. “There was also little clarity about what to expect from the exam.”

Since its launch, TikuretEntrance has onboarded more than 200 active users and was part of a startup mentorship program under the Disruptor’s Den, a platform for emerging Ethiopian entrepreneurs.

Through a personalized dashboard, users can access previous entrance exams, build customized study schedules, follow guided learning paths and track progress over time. Performance analytics highlight strengths and gaps across subjects, while an exam-countdown tool helps students plan their preparation more strategically.

The platform also includes study groups and peer-support features, allowing students to revise collaboratively, alongside a growing digital library of learning resources that is updated as curricula change. Additional tools offer revision strategies, preparation tips and early access to new features.

Unlike platforms that emphasize large volumes of downloadable content, TikuretEntrance focuses on consistency and habit-building, encouraging students to study in smaller, regular intervals rather than cramming at the last minute.

To support learners outside school hours, the company offers round-the-clock assistance through WhatsApp and Telegram, where students can ask questions, report technical problems and receive academic guidance. The service is fully mobile-friendly and runs through a web interface rather than a dedicated app, a choice intended to reduce barriers for students using lower-end phones or devices with limited storage.

TikuretEntrance operates on a subscription model, charging 499 Birr for annual access, and is pursuing partnerships with secondary schools as well as referral programs to expand adoption in both public and private institutions.

For now, the platform is ce

ntered on individual learners. But the founders say they are developing tools that would allow parents and schools to monitor student progress, a sign of growing interest in data-driven approaches to exam preparation in a system where the stakes of a single test have rarely felt higher.

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Daniel Metaferiya

Daniel Metaferiya is a writer, journalist and radio host, with a keen interest in technology. He follows developments in Ethiopia's startup ecosystem closely and is passionate about profiling unique MSMEs.

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For grade eleventh chemistry final exam answers

For grade eleventh chemistry final exam answers
Grade 11th Chemistry Final Examination in Jan, 2026
Answer Sheet
Name: __ Grade & Section: __

I. TRUE/FALSE
1. True

False

True

False

False

True

True

False

False

True

II. Matching
C

B

E

A

D

III. Multiple Choice
C

C

C

C

C

A

B

B

A

A

A

B

B

B

C

C

B

D

D

B

IV. Fill the Blank Spaces
Dipole

3

[He] 2s² 2p² (Orbital Diagram below)

Formation of Lattice Energy (or Electron Affinity)

40.Photoelectric effect

Grade ten first semester chemistry final exam answers

Grade eleventh chemistry
Hybridization

Grade eleventh chemistry
https://youtu.be/nNbEi5Yul6E?si=Xj8IZiyICVdx8Y-x

– C) sp³
    – D) dsp³

▎Review Questions on Bonding Theories

26. Which type of bond is formed by the lateral overlap of p orbitals?
    – A) Sigma bond
    – B) Pi bond
    – C) Delta bond
    – D) Ionic bond

27. Which hybridization corresponds to a tetrahedral geometry?
    – A) sp
    – B) sp²
    – C) sp³
    – D) dsp³

28. The presence of lone pairs affects the geometry of a molecule by:
    – A) Increasing bond angles.
    – B) Decreasing bond angles.
    – C) Not affecting bond angles.
    – D) Creating new bonds.

29. In which molecule does the central atom exhibit dsp² hybridization?
    – A) XeF₄
    – B) BF₃
    – C) NH₃
    – D) CH₄

30. The total number of orbitals involved in forming a triple bond is:
    – A) 1 sigma and 1 pi
    – B) 1 sigma and 2 pi
    – C) 3 sigma and no pi
    – D) No sigma and 3 pi

eleventh-grade chemistry:

▎Electron Configuration of Diatomic Molecules

1. Which of the following diatomic molecules has the electron configuration of 1s² 2s² 2p²?
– A) O₂
– B) C₂
– C) N₂
– D) F₂

2. What is the bond order of the molecule O₂?
– A) 1
– B) 2
– C) 3
– D) 4

3. Which of the following molecules is paramagnetic?
– A) N₂
– B) O₂
– C) F₂
– D) He₂

4. The electron configuration for N₂ is:
– A) 1s² 2s² 2p³
– B) 1s² 2s² 2p⁴
– C) 1s² 2s² 2p⁵
– D) 1s² 2s² 2p⁶

5. Which of the following statements is true about diamagnetic substances?
– A) They have unpaired electrons.
– B) They are attracted to a magnetic field.
– C) They have all paired electrons.
– D) They exhibit strong magnetism.

▎Chemical Bonding Theories

6. Which theory explains the formation of covalent bonds through the overlap of atomic orbitals?
– A) VSEPR Theory
– B) Valence Bond Theory
– C) Molecular Orbital Theory
– D) Ionic Bonding Theory

7. In molecular orbital theory, which type of bond results from the head-on overlap of orbitals?
– A) Pi bond
– B) Sigma bond
– C) Delta bond
– D) Gamma bond

8. What type of hybridization occurs in CH₄?
– A) sp
– B) sp²
– C) sp³
– D) sp³d

9. How many sigma and pi bonds are present in the molecule C₂H₄?
– A) 6 sigma, 0 pi
– B) 4 sigma, 2 pi
– C) 5 sigma, 1 pi
– D) 3 sigma, 3 pi

10. The hybridization of the central atom in NH₃ is:
▪ A) sp
▪ B) sp²
▪ C) sp³
▪ D) dsp²

▎Types of Overlapping of Orbitals

11. What type of overlap occurs when two p orbitals form a pi bond?
– A) Side-to-side overlap
– B) End-to-end overlap
– C) Hybrid overlap
– D) None of the above

12. In which molecule is a double bond present?
– A) H₂
– B) N₂
– C) O₂
– D) Cl₂

13. Which type of bond involves the sharing of one pair of electrons?
– A) Triple bond
– B) Double bond
– C) Single bond
– D) Quadruple bond

14. In a molecule with sp hybridization, how many sigma bonds can be formed?
– A) 1
– B) 2
– C) 3
– D) 4

15. The total number of sigma bonds in a molecule with a triple bond and two single bonds is:
– A) 3
– B) 4
– C) 5
– D) 6

▎Hybridization and Bonding Characteristics

16. Which of the following molecules exhibits sp² hybridization?
– A) C₂H₆
– B) C₂H₄
– C) C₂H₂
– D) CH₄

17. The bond angle in an sp hybridized molecule is approximately:
– A) 90°
– B) 120°
– C) 180°
– D) 109.5°

18. What is the hybridization of the central atom in SF₆?
– A) sp
– B) sp²
– C) sp³
– D) sp³d²

19. Which molecule has a linear geometry due to its hybridization?
– A) CO₂
– B) H₂O
– C) NH₃
– D) CH₄

20. The bond order for a molecule with two sigma bonds and one pi bond is:
– A) 1
– B) 1.5
– C) 2
– D) 3

▎Advanced Concepts in Bonding

21. Which of the following statements about molecular orbital theory is incorrect?
– A) Electrons in molecular orbitals are delocalized.
– B) Molecular orbitals can be bonding or antibonding.
– C) Molecular orbital theory only applies to diatomic molecules.
– D) Bond order can be calculated using molecular orbital theory.

22. In molecular orbital theory, which orbital is higher in energy than the 3s?
– A) 3p
– B) 3d
– C) 4s
– D) All of the above

23. The molecule F₂ has how many unpaired electrons in its ground state?
– A) 0
– B) 1
– C) 2
– D) 3

24. If a molecule has a bond order of zero, it means:
– A) It is stable.
– B) It cannot exist.
– C) It has no bonds.
– D) It has equal numbers of bonding and antibonding electrons.

25. The hybridization for a molecule with four regions of electron density around a central atom is:
– A) sp
– B) sp²

Nature
https://www.nature.com
Biogenic synthesis, characterization, and antibacterial ...
https://www.google.com/url?sa=t&source=web&rct=j&opi=89978449&url=https://www.nature.com/articles/s41598-026-37265-9&ved=2ahUKEwio3PC_1_qSAxVlwAIHHV7lIGAQFnoECBcQAQ&usg=AOvVaw3FrwmSSW9y9S3piRivu972

Grades eleventh

Grade ten acids

Grade ten unit three 3.4 base